compound assignment operators maze

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1.5. Compound Assignment Operators ¶

Compound assignment operators are shortcuts that do a math operation and assignment in one step. For example, x += 1 adds 1 to x and assigns the sum to x. It is the same as x = x + 1 . This pattern is possible with any operator put in front of the = sign, as seen below.

The most common shortcut operator ++ , the plus-plus or increment operator, is used to add 1 to the current value; x++ is the same as x += 1 and the same as x = x + 1 . It is a shortcut that is used a lot in loops. If you’ve heard of the programming language C++, the ++ in C++ is an inside joke that C has been incremented or improved to create C++. The -- decrement operator is used to subtract 1 from the current value: y-- is the same as y = y - 1 . These are the only two double operators; this shortcut pattern does not exist with other operators. Run the following code to see these shortcut operators in action!

Run the code below to see what the ++ and shorcut operators do. Use the Codelens to trace through the code and observe how the variable values change. Try creating more compound assignment statements with shortcut operators and guess what they would print out before running the code.

1-5-2: What are the values of x, y, and z after the following code executes?

• x = -1, y = 1, z = 4
• This code subtracts one from x, adds one to y, and then sets z to to the value in z plus the current value of y.
• x = -1, y = 2, z = 3
• x = -1, y = 2, z = 2
• x = -1, y = 2, z = 4

1-5-3: What are the values of x, y, and z after the following code executes?

• x = 6, y = 2.5, z = 2
• This code sets x to z * 2 (4), y to y divided by 2 (5 / 2 = 2) and z = to z + 1 (2 + 1 = 3).
• x = 4, y = 2.5, z = 2
• x = 6, y = 2, z = 3
• x = 4, y = 2.5, z = 3
• x = 4, y = 2, z = 3

1.5.1. Code Tracing Challenge and Operators Maze ¶

Code Tracing is a technique used to simulate by hand a dry run through the code or pseudocode as if you are the computer executing the code. Tracing can be used for debugging or proving that your program runs correctly or for figuring out what the code actually does.

Trace tables can be used to track the values of variables as they change throughout a program. To trace through code, write down a variable in each column or row in a table and keep track of its value throughout the program. Some trace tables also keep track of the output and the line number you are currently tracing.

For example, given the following code:

The corresponding trace table looks like this:

Alternatively, we can show a compressed trace by listing the sequence of values assigned to each variable as the program executes. You might want to cross off the previous value when you assign a new value to a variable. The last value listed is the variable’s final value.

Use paper and pencil to trace through the following program to determine the values of the variables at the end. Be careful, % is the remainder operator, not division.

1.5.2. Prefix versus Postfix Operator ¶

What do you think is printed when the following code is executed? Try to guess the output before running the code. You might be surprised at the result. Click on CodeLens to step through the execution. Notice that the second println prints the original value 7 even though the memory location for variable count is updated to the value 8.

The code System.out.println(count++) adds one to the variable after the value is printed. Try changing the code to ++count and run it again. This will result in one being added to the variable before its value is printed. When the ++ operator is placed before the variable, it is called prefix increment. When it is placed after, it is called postfix increment.

• System.out.println(score++);
• Print the value 5, then assign score the value 6.
• System.out.println(score--);
• Print the value 5, then assign score the value 4.
• System.out.println(++score);
• Assign score the value 6, then print the value 6.
• System.out.println(--score);
• Assign score the value 4, then print the value 4.

When you are new to programming, it is advisable to avoid mixing unary operators ++ and -- with assignment or print statements. Try to perform the increment or decrement operation on a separate line of code from assignment or printing.

For example, instead of writing x=y++; or System.out.println(z--); the code below makes it clear that the increment of y happens after the assignment to x , and that the value of z is printed before it is decremented.

• System.out.println(score); score++;
• System.out.println(score); score--;
• score++; System.out.println(score);
• score--; System.out.println(score);

1.5.3. Summary ¶

Compound assignment operators (+=, -=, *=, /=, %=) can be used in place of the assignment operator.

The increment operator (++) and decrement operator (–) are used to add 1 or subtract 1 from the stored value of a variable. The new value is assigned to the variable.

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1.5. Compound Assignment Operators ¶

Compound assignment operators are shortcuts that do a math operation and assignment in one step. For example, x += 1 adds 1 to the current value of x and assigns the result back to x . It is the same as x = x + 1 . This pattern is possible with any operator put in front of the = sign, as seen below. If you need a mnemonic to remember whether the compound operators are written like += or =+ , just remember that the operation ( + ) is done first to produce the new value which is then assigned ( = ) back to the variable. So it’s operator then equal sign: += .

Since changing the value of a variable by one is especially common, there are two extra concise operators ++ and -- , also called the plus-plus or increment operator and minus-minus or decrement operator that set a variable to one greater or less than its current value.

Thus x++ is even more concise way to write x = x + 1 than the compound operator x += 1 . You’ll see this shortcut used a lot in loops when we get to them in Unit 4. Similarly, y-- is a more concise way to write y = y - 1 . These shortcuts only exist for + and - as they don’t really make sense for other operators.

If you’ve heard of the programming language C++, the name is an inside joke that C, an earlier language which C++ is based on, had been incremented or improved to create C++.

Here’s a table of all the compound arithmetic operators and the extra concise incremend and decrement operators and how they relate to fully written out assignment expressions. You can run the code below the table to see these shortcut operators in action!

Run the code below to see what the ++ and shorcut operators do. Click on the Show Code Lens button to trace through the code and the variable values change in the visualizer. Try creating more compound assignment statements with shortcut operators and work with a partner to guess what they would print out before running the code.

If you look at real-world Java code, you may occassionally see the ++ and -- operators used before the name of the variable, like ++x rather than x++ . That is legal but not something that you will see on the AP exam.

Putting the operator before or after the variable only changes the value of the expression itself. If x is 10 and we write, System.out.println(x++) it will print 10 but aftewards x will be 11. On the other hand if we write, System.out.println(++x) , it will print 11 and afterwards the value will be 11.

In other words, with the operator after the variable name, (called the postfix operator) the value of the variable is changed after evaluating the variable to get its value. And with the operator before the variable (the prefix operator) the value of the variable in incremented before the variable is evaluated to get the value of the expression.

But the value of x after the expression is evaluated is the same in either case: one greater than what it was before. The -- operator works similarly.

The AP exam will never use the prefix form of these operators nor will it use the postfix operators in a context where the value of the expression matters.

1-5-2: What are the values of x, y, and z after the following code executes?

• x = -1, y = 1, z = 4
• This code subtracts one from x, adds one to y, and then sets z to to the value in z plus the current value of y.
• x = -1, y = 2, z = 3
• x = -1, y = 2, z = 2
• x = 0, y = 1, z = 2
• x = -1, y = 2, z = 4

1-5-3: What are the values of x, y, and z after the following code executes?

• x = 6, y = 2.5, z = 2
• This code sets x to z * 2 (4), y to y divided by 2 (5 / 2 = 2) and z = to z + 1 (2 + 1 = 3).
• x = 4, y = 2.5, z = 2
• x = 6, y = 2, z = 3
• x = 4, y = 2.5, z = 3
• x = 4, y = 2, z = 3

1.5.1. Code Tracing Challenge and Operators Maze ¶

Use paper and pencil or the question response area below to trace through the following program to determine the values of the variables at the end.

Code Tracing is a technique used to simulate a dry run through the code or pseudocode line by line by hand as if you are the computer executing the code. Tracing can be used for debugging or proving that your program runs correctly or for figuring out what the code actually does.

Trace tables can be used to track the values of variables as they change throughout a program. To trace through code, write down a variable in each column or row in a table and keep track of its value throughout the program. Some trace tables also keep track of the output and the line number you are currently tracing.

Trace through the following code:

1-5-4: Write your trace table for x, y, and z here showing their results after each line of code.

After doing this challenge, play the Operators Maze game . See if you and your partner can get the highest score!

1.5.2. Summary ¶

Compound assignment operators ( += , -= , *= , /= , %= ) can be used in place of the assignment operator.

The increment operator ( ++ ) and decrement operator ( -- ) are used to add 1 or subtract 1 from the stored value of a variable. The new value is assigned to the variable.

The use of increment and decrement operators in prefix form (e.g., ++x ) and inside other expressions (i.e., arr[x++] ) is outside the scope of this course and the AP Exam.

Assign2: Mazes

A maze is a twisty and convoluted arrangement of paths that challenge the solver to find a route from the entry to the exit. This assignment is about using ADTs to represent, process, and solve mazes.

Labyrinths and mazes have fascinated humans since ancient times (remember Theseus and the Minotaur? ), but mazes can be more than just recreation. The mathematician Leonhard Euler was one of the first to analyze plane mazes mathematically, and in doing so founded the branch of mathematics known as topology. Many algorithms that operate on mazes are closely related to graph theory and have applications to diverse tasks such as designing circuit boards, routing network traffic, motion planning, and social networking.

This project focuses on a particular type of maze known as a perfect maze. A perfect, or simply-connected, maze has no loops and no inaccessible areas. Any two locations within a perfect maze are connected by exactly one path.

The image below shows a perfect maze and the path from the entry at upper left to exit at lower right:

Grid and GridLocation

A maze can be modeled as a two-dimensional array where each element is either a wall or a corridor. The Grid class from the Stanford library is a good fit tool for representing a maze. It provides the abstraction of a two-dimensional array in a safe, encapsulated form with various client conveniences. We use a bool as the element, storing true for a cell that is an open corridor and false for a wall.

Access to grid elements is generally by row and column, e.g. grid [ row ][ col ] . Alternatively elements can be selected by GridLocation . GridLocation is a small struct type that pairs the row and column together into one aggregate type. A struct is a C++ user-defined type consisting of a defined set of named fields of heterogeneous type.

• Lecture slides on Grid
• Documentation for Grid
• Section 6.1 of the textbook introduces struct types
• Header files grid . h and gridlocation . h within the starter project subfolder lib / StanfordCPPLIb / collections /
• GridLocation has a neighbors operation that is a bit ill-matched to our needs. For maze, neighbors are defined in the four cardinal directions only (N,S,E,W), but GridLocation also considers NE, NW and so on. Rather than try to work around the mismatch, it is cleaner to write your own operation to collect the neighbors.

Reading a maze file

A maze file contains a maze written in text format. Each line of the file corresponds to one row of the maze. Within a row, the character @ is used for walls and - for corridors. Here are the first few lines of a sample maze file:

The last line of a maze file must either be a blank line or can optionally includes the solution written as a Stack of GridLocation.

A completed version of the function

is provided in the starter project. The provided function correctly reads a well-formed maze, which is confirmed by the provided unit tests. First read over the function and its unit tests. Ask questions if you find anything unclear.

Although readMazeFile does a great job reading correct files, it does a poor job if asked to read a malformed file. Since it makes no effort to detect problems, it generally blunders through, misinterpreting the data, and possibly crashing on bad inputs.

Add error detection

Your first task is to improve readMazeFile by making it robust to malformed input. There are two specific issues in the file format that you are to detect and report:

• Each maze row is required to have same length as all the others.
• The valid options for a location are open or wall. Anything else is an error.

Use the error function from the header "error.h" to report an error. This function is used to report a fatal error. When you call error it halts the program right here and reports the message you provided:

Construct tests and maze files

We’ve provided some unit tests and one badly malformed maze file. You are to add student tests that further verify that your maze reading is now robust. You will need to create additional malformed files. Maze files are just ordinary text files placed into the res / folder of your project. You can edit these files in Qt Creator. The project resource files are listed under Other files -> res . Your program can open a resource file by specifying the path to open as "res/myfilename" . Follow the format of the other maze files: one line for each row in grid, followed by one extra line that is blank or the solution path written as a Stack of GridLocation .

Note the use of the special kind of test case EXPECT_ERROR in our provided test. An EXPECT_ERROR test case evaluates an expression, expected that the operation with raise an error. While the test is running, SimpleTest framework will catch the error, note that it was generated, and then resume. Because the error that was raised was expected, the test case is considered to be passed. If an error was expected but didn't materialize, the test case fails. (Note this is opposite of the regular EXPECT and EXPECT_EQUAL tests which do not expect errors and treat a raised error as a test failure). More information on the different test macros and how they all work can be found in the CS106B Testing Guide

• Q6. Describe the malformed maze files you created in order to verify the robustness of your maze reading.

Check solution

The last line of a maze file must either be a blank line or can optionally includes the solution written as a Stack of GridLocation. The readMazeFile function includes the correct code to read the Stack and validate it is syntactically well-formed. To confirm that the path is a valid solution, it calls the function checkSolution .

The function

is intended to verify that a path meets all of the necessary criteria to be a correct solution:

• The path must start at the entry (upper left corner).
• The path must end at the exit (lower right corner).
• Each location in the path is within the maze bounds.
• Each location in the path is an open corridor (not wall).
• Each location is one cardinal step (N,S,E,W) from the next in path.
• The path contains no loops, i.e. a location appears at most once in the path.

If checkSolution detects any of the above problems, it should call the error function with an appropriate message. A call to error immediately exits the function and halts the program. If all of the criteria are met, then the function completes normally and returns true to signal success.

This function has a lot of things to check! However, writing this function now is going to pay off big time when you later use it to test your path-finding algorithm. Be sure to very thoroughly test your checkSolution now on a variety of invalid paths so that you can be confident it is the oracle of truth when it comes to validating a solution. Your future self will thank you.

• Q7. After you have written your tests, describe your testing strategy to determine that your checkSolution works as intended

Once your maze-reading is bullet-proofed against all manner of problems, you're ready to go on to display the maze and code the algorithm to solve it.

Drawing the maze

The Stanford C++ library has extensive functionality for drawing and interacting with the user, but we generally don't ask you to dig into those features. Instead, we have supplied the graphics routines you need pre-written in a simple, easy-to-use form. For this program, the provided mazegraphics . cpp module has functions to display a maze and highlight a path through the maze. Read the mazegraphics . h header file in the starter project for details of these functions:

• MazeGraphics :: drawGrid ( Grid < bool >& grid )
• MazeGraphics :: highlightPath ( Stack < GridLocation > path , string color )

After reading in a maze (and validating that it is well-formed), you simply call the MazeGraphics :: drawGrid to display the maze. The function MazeGraphics :: highlightPath is later used to mark the cells along a path as part of animating the search for a solution. Highlighting a path expects to add dots on locations within the already drawn grid. Be sure to only call MazeGraphics :: highlightPath after having first called MazeGraphics :: drawGrid once.

Note that when calling these functions, you must call them using their full name (including the weird looking MazeGraphics :: prefix). We will talk more about what this notation means a little later on in the quarter!

Solving a maze using breadth-first search (BFS)

Now that we know we have the ability to construct a maze, we can turn our attention to figuring out how we will escape these mazes we've just built. In particular, you are now going to implement the following function in maze . cpp , in which you will implement a maze-solving algorithm (described below).

There are a wide variety of algorithms for solving a maze. Solving a maze can be seen as a specific instance of a shortest path problem, where the challenge is to find the shortest route from the entrance to the exit. Shortest path problems come up in a variety of situations such as packet routing, robot motion planning, analyzing gene mutations, spell correction, and more. In the case of the perfect maze, the shortest path is also the only path, but the general process is the same regardless.

Breadth-first search (BFS) is a classic and elegant algorithm for finding a shortest path. A breadth-first search reaches outward from the entry location in a radial fashion until it finds the exit. The first paths examined take one hop from the entry. If any of these reach the exit location, you’re done. If not, the search expands to those paths that are two hops long. At each subsequent step, the search expands radially, examining all paths of length three, then of length four, etc.), stopping at the first path that reaches the exit.

Breadth-first search is typically implemented using a queue. The queue stores partial paths that represent possibilities to explore. The paths are processed in order of increasing length. The first paths enqueued are all length one, followed by the length two paths, and so on. Given the FIFO handling of the queue, all shorter paths are dequeued before the longer paths make their way to the front of queue, ready for their turn to be processed.

At each step, the algorithm considers the current path frontmost in the queue. If the current path ends at the exit, it is a complete solution. If not, the algorithm takes the current path and extends it to reach locations that are one hop further away in the maze, and enqueue those extended paths to be examined later.

To represent a path, a Stack of GridLocation is a good choice, as it cleanly supports the needed operations to add/examine the element at the end of the path. Putting these paths into a queue for processing means you'll have a nested ADT, a Queue < Stack < GridLocation >> . A nested container type looks a little scary at first, but it is just the right tool for this job.

Here are steps followed by a breadth-first search:

• Create a queue of paths. A path is a stack of grid locations.
• For simplicity, assume entry is always the upper-left corner and exit in the lower-right.
• Dequeue path from queue.
• If this path ends at exit, this path is the solution!
• For each viable neighbor of path end, make copy of path, extend by adding neighbor and enqueue it.
• A location has up to four neighbors, one in each of the four cardinal directions. A neighbor location is viable if it is within the maze bounds, the cell is an open corridor (not a wall), and it has not yet been visited.
• Repeat steps 3-5 until path reaches exit.

One issue that is a bit subtle is that you must avoid repeatedly revisiting the same location in the maze or creating a path with a cycle, lest the search get stuck in a infinite loop. For example, if the current path leads from location r0c0 to r1c0 , you should not extend the path by moving back to location r0c0 . A common strategy for tracking where you have already been is to keep a Set of locations to which you add each location as you visit it. Checking whether a location is contained in the visited set lets you know whether to consider or skip this location.

As your algorithm hunts for a solution, you can call the function MazeGraphics :: highlightPath function to visually mark the current path on the graphics window so the user can follow along with the algorithm. This animation is not required, but it adds a bit of fun and may even help you trace and debug.

Once you have a working solver, unit-testing is a piece of cake because earlier you wrote the awesome checkSolution function. After solving a maze, use your unit test verify that your path meets all the criteria to be valid. Super neat! Your work here is done, congratulations!👏

Notes on Collection ADTs

• For solving the maze, the Stack with its LIFO behavior is ideal for storing a path. The first entry pushed on the stack is the starting point and each subsequent point in the path is pushed on top. The FIFO Queue is just what's needed to track those partial paths under consideration. The paths are enqueued (and thus dequeued) in order of increasing length. A Set is particularly handy if you need to be able to quickly determine membership (is this element contained in the set?).
• The assignment operator work as expected for all our ADTs. Assigning from one Stack/Vector/Set/etc to another will create a copy of the ADT with a copy of the elements.
• Be on your toes about making sure that your template types are always properly specialized and that all types match. Using Vector without specifying the element type just won't fly, and a Vector < int > is not the same thing as a Vector < double > . The error messages you receive when you have mismatches can be cryptic and hard to interpret. Bring your template woes to the forum and we can help untangle them with you.

There are many interesting facets to mazes and much fascinating mathematics underlying them. Mazes will come up again several times this quarter. Chapter 9 of the textbook uses a recursive depth-first search as path-finding algorithm. At the end of the quarter when we talk about graphs, we'll explore the equivalence between mazes and graphs and note how many of the interesting results in mazes are actually based on work done in terms of graph theory.

• Walter Pullen, Maze Classification . http://www.astrolog.org/labyrnth/algrithm.htm Website with lots of great info on mazes and maze algorithms
• Jamis Buck. Maze Algorithms . https://www.jamisbuck.org/mazes/ Fun animations of maze algorithms. He also wrote the excellent book about Mazes for Programmers: Code Your Own Twisty Little Passages .
• Instead of reading pre-written mazes from a file, you could instead generate a new random maze on demand. There is an amazing (I could not resist…) variety of algorithms for maze construction, ranging from the simple to the sublime. Here are a few names to get you started: backtracking, depth-first, growing tree, sidewinder, along with algorithms named for their inventor: Aldous-Broder, Eller, Prim, Kruskal, Wilson, and many others.
• Try out other maze solving algorithms. How does BFS stack up against the options in terms of solving power or runtime efficiency? Take a look at random mouse, wall following, depth-first search (either manually using a Stack or using recursion once you get the hang of it), Bellman-Ford, or others.
• There are many other neat maze-based games out there that make fun extensions. You might gather ideas from Robert Abbott's http://www.logicmazes.com or design a maze game board after inspiration from Chris Smith on mazes .

© Stanford 2020 · Website by Julie Zelenski · Page updated 2020-Apr-23

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AP Computer Science A

Compound Assignment Operators

Compound assignment operators are shorthand notations that combine an arithmetic operation with the assignment operator. They allow you to perform an operation and assign the result to a variable in a single step.

Related terms

+= operator : This operator adds the value on the right side of it to the variable on the left side, and then assigns the result back to that variable.

-= operator : This operator subtracts the value on the right side of it from the variable on the left side, and then assigns the result back to that variable.

*= operator : This operator multiplies the value on the right side of it with the variable on the left side, and then assigns the result back to that variable.

" Compound Assignment Operators " appears in:

Study guides ( 1 ).

• AP Computer Science A - 1.4 Compound Assignment Operators

Practice Questions ( 1 )

• Which arithmetic operators do not have compound assignment operators?

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Java Compound Operators

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• Java Operators

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1. Overview

In this tutorial, we’ll have a look at Java compound operators, their types and how Java evaluates them.

We’ll also explain how implicit casting works.

2. Compound Assignment Operators

An assignment operator is a binary operator that assigns the result of the right-hand side to the variable on the left-hand side. The simplest is the “=” assignment operator:

This statement declares a new variable x , assigns x the value of 5 and returns 5 .

Compound Assignment Operators are a shorter way to apply an arithmetic or bitwise operation and to assign the value of the operation to the variable on the left-hand side.

For example, the following two multiplication statements are equivalent, meaning  a and b will have the same value:

It’s important to note that the variable on the left-hand of a compound assignment operator must be already declared. In other words,  compound operators can’t be used to declare a new variable.

Like the “=” assignment operator, compound operators return the assigned result of the expression:

Both x and y will hold the value 3 .

The assignment (x+=2) does two things: first, it adds 2 to the value of the variable x , which becomes  3;  second, it returns the value of the assignment, which is also 3 .

3. Types of Compound Assignment Operators

Java supports 11 compound assignment operators. We can group these into arithmetic and bitwise operators.

Let’s go through the arithmetic operators and the operations they perform:

• Incrementation: +=
• Decrementation: -=
• Multiplication: *=
• Division: /=
• Modulus: %=

Then, we also have the bitwise operators:

• AND, binary: &=
• Exclusive OR, binary: ^=
• Inclusive OR, binary: |=
• Left Shift, binary: <<=
• Right Shift, binary: >>=
• Shift right zero fill: >>>=

Let’s have a look at a few examples of these operations:

As we can see here, the syntax to use these operators is consistent.

4. Evaluation of Compound Assignment Operations

There are two ways Java evaluates the compound operations.

First, when the left-hand operand is not an array, then Java will, in order:

• Verify the operand is a declared variable
• Save the value of the left-hand operand
• Evaluate the right-hand operand
• Perform the binary operation as indicated by the compound operator
• Convert the result of the binary operation to the type of the left-hand variable (implicit casting)
• Assign the converted result to the left-hand variable

Next, when the left-hand operand is an array, the steps to follow are a bit different:

• Verify the array expression on the left-hand side and throw a NullPointerException  or  ArrayIndexOutOfBoundsException if it’s incorrect
• Save the array element in the index
• Check if the array component selected is a primitive type or reference type and then continue with the same steps as the first list, as if the left-hand operand is a variable.

If any step of the evaluation fails, Java doesn’t continue to perform the following steps.

Let’s give some examples related to the evaluation of these operations to an array element:

As we’d expect, this will throw a  NullPointerException .

However, if we assign an initial value to the array:

We would get rid of the NullPointerException, but we’d still get an  ArrayIndexOutOfBoundsException , as the index used is not correct.

If we fix that, the operation will be completed successfully:

Finally, the x variable will be 6 at the end of the assignment.

5. Implicit Casting

One of the reasons compound operators are useful is that not only they provide a shorter way for operations, but also implicitly cast variables.

Formally, a compound assignment expression of the form:

is equivalent to:

E1 – (T)(E1 op E2)

where T is the type of E1 .

Let’s consider the following example:

Let’s review why the last line won’t compile.

Java automatically promotes smaller data types to larger data ones, when they are together in an operation, but will throw an error when trying to convert from larger to smaller types .

So, first,  i will be promoted to long and then the multiplication will give the result 10L. The long result would be assigned to i , which is an int , and this will throw an error.

This could be fixed with an explicit cast:

Java compound assignment operators are perfect in this case because they do an implicit casting:

This statement works just fine, casting the multiplication result to int and assigning the value to the left-hand side variable, i .

6. Conclusion

In this article, we looked at compound operators in Java, giving some examples and different types of them. We explained how Java evaluates these operations.

Finally, we also reviewed implicit casting, one of the reasons these shorthand operators are useful.

As always, all of the code snippets mentioned in this article can be found in our GitHub repository .

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• Module 1 Mixed Up Code Practice

Compound Assignment Operators ¶

Compound assignment operators are shortcuts that do a math operation and assignment in one step. For example, x += 1 adds 1 to x and assigns the sum to x. It is the same as x = x + 1 . This pattern is possible with any operator put in front of the = sign, as seen below.

The most common shortcut operator ++ , the plus-plus or increment operator, is used to add 1 to the current value; x++ is the same as x += 1 and the same as x = x + 1 . It is a shortcut that is used a lot in loops. If you’ve heard of the programming language C++, the ++ in C++ is an inside joke that C has been incremented or improved to create C++. The -- decrement operator is used to subtract 1 from the current value: y-- is the same as y = y - 1 . These are the only two double operators; this shortcut pattern does not exist with other operators. Run the following code to see these shortcut operators in action!

Run the code in E01ShortcutOperators to see what the ++ and shorcut operators do. Use the debugger to trace through the code and observe how the variable values change. Try creating more compound assignment statements with shortcut operators and guess what they would print out before running the code.

1-5-1: What are the values of x, y, and z after the following code executes?

• x = -1, y = 1, z = 4
• This code subtracts one from x, adds one to y, and then sets z to to the value in z plus the current value of y.
• x = -1, y = 2, z = 3
• x = -1, y = 2, z = 2
• x = -1, y = 2, z = 4

1-5-2: What are the values of x, y, and z after the following code executes?

• x = 6, y = 2.5, z = 2
• This code sets x to z * 2 (4), y to y divided by 2 (5 / 2 = 2) and z = to z + 1 (2 + 1 = 3).
• x = 4, y = 2.5, z = 2
• x = 6, y = 2, z = 3
• x = 4, y = 2.5, z = 3
• x = 4, y = 2, z = 3

Code Tracing Challenge and Operators Maze ¶

Code Tracing is a technique used to simulate by hand a dry run through the code or pseudocode as if you are the computer executing the code. Tracing can be used for debugging or proving that your program runs correctly or for figuring out what the code actually does.

Trace tables can be used to track the values of variables as they change throughout a program. To trace through code, write down a variable in each column or row in a table and keep track of its value throughout the program. Some trace tables also keep track of the output and the line number you are currently tracing.

For example, given the following code:

The corresponding trace table looks like this:

Alternatively, we can show a compressed trace by listing the sequence of values assigned to each variable as the program executes. You might want to cross off the previous value when you assign a new value to a variable. The last value listed is the variable’s final value.

Use paper and pencil to trace through the following program to determine the values of the variables at the end. Be careful, % is the remainder operator, not division.

The final value for x is

The final value for y is

The final value for z is

Prefix versus Postfix Operator ¶

Open the E02PostfixExample program. What do you think is printed when the following code is executed? Try to guess the output before running the code. You might be surprised at the result. Use the debugger to step through the execution. Notice that the second println prints the original value 7 even though the memory location for variable count is updated to the value 8.

The code System.out.println(count++) adds one to the variable after the value is printed. Try changing the code to ++count and run it again. This will result in one being added to the variable before its value is printed. When the ++ operator is placed before the variable, it is called prefix increment. When it is placed after, it is called postfix increment.

• System.out.println(score++);
• Print the value 5, then assign score the value 6.
• System.out.println(score--);
• Print the value 5, then assign score the value 4.
• System.out.println(++score);
• Assign score the value 6, then print the value 6.
• System.out.println(--score);
• Assign score the value 4, then print the value 4.

When you are new to programming, it is advisable to avoid mixing unary operators ++ and -- with assignment or print statements. Try to perform the increment or decrement operation on a separate line of code from assignment or printing.

For example, instead of writing x=y++; or System.out.println(z--); the code below makes it clear that the increment of y happens after the assignment to x , and that the value of z is printed before it is decremented.

• System.out.println(score); score++;
• System.out.println(score); score--;
• score++; System.out.println(score);
• score--; System.out.println(score);

Compound assignment operators (+=, -=, *=, /=, %=) can be used in place of the assignment operator.

The increment operator ( ++ ) and decrement operator ( – ) are used to add 1 or subtract 1 from the stored value of a variable. The new value is assigned to the variable.

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Computer Science A

Compound Assignment Operators

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Compound Assignment Operators: AP Computer Science A Study Guide

Introduction.

Welcome, aspiring coders and budding programmers, to the world of compound assignment operators in Java! Get ready to level up your coding skills by learning these shorthand operations that will make your code both snappier and cleaner. Plus, stick around for some fun examples, jokes, and puns that will keep you entertained while you learn. 🚀💻

Understanding Compound Assignment Operators

Imagine you have a variable that's a bit like a snowball—every time you roll it down a hill, it picks up more snow and gets bigger. Now, picture typing out "roll the snowball" every time that happens. It's kind of repetitive, right? Enter compound assignment operators: they're your shortcut to making things more efficient.

Let's start with a basic example:

This could be snazzier, like this:

It's like saying, "Here's 3 more snowflakes for you, snowball!" The += is a compound assignment operator that takes the current value of snowball , adds 3 to it, and sets that as the new value. Elementary, my dear Watson!

But wait, there are more! Here are the other compound assignment operators you can use:

• += for addition: score += 10; 🎮
• -= for subtraction: inventory -= 2; 🪓
• *= for multiplication: distance *= 2; 🏃
• /= for division: fuel /= 2; ⛽
• %= for modulus (remainder): turns %= 3; 🌀

Incrementing and Decrementing

Alright, programmers, let's talk about adding and subtracting one, because who doesn't love a good +1 or -1?

Consider these lines:

Or, even simpler:

When you want to add 1 ( ++ ) or subtract 1 ( -- ), there's an easier way. You can use pre-increment (++i) or post-increment (i++). It's kind of like picking your favorite Pokémon starter—you've got choices!

Let's break it down with a friendly duel:

In the first line, xp++ prints the current xp before increasing it. In the second line, ++xp increases the xp first before printing.

To visualize:

• xp++ is like saying, "Hey, use xp now and then give it a boost."
• ++xp is like saying, "Give xp a boost, then use it."

Code Tracing Practice 🕵️‍♂️

Time to put on your detective hat and become a code detective! 🕵️‍♀️ Code tracing means following each line of code to see what our variables are up to. Ready for some practice?

Step-by-step how it works:

• potion *= 3; --> potion is now 18.
• ingredient -= 2; --> ingredient is now 2.
• result = potion % ingredient; --> result is now 0.
• potion += result; --> potion remains 18.
• ingredient = potion - ingredient; --> ingredient is now 16.
• result *= ingredient; --> result remains 0.

Final values: potion = 18, ingredient = 16, result = 0.

Let's follow these enchanted lines:

• x /= y; --> x is now 3.75.
• y *= x; --> y is now 15.0.
• z = y % x; --> z is now 3.75.
• x += z; --> x is now 7.5.
• y = x / z; --> y is now 2.0.
• z *= y; --> z is now 7.5.

Final values: x = 7.5, y = 2.0, z = 7.5.

Feel free to practice more on your own or with a friend. The more you trace, the better you get at it! Plus, we've got a fun little challenge for you: try out the Operators Maze game by CSAwesome!

Key Terms to Review

To wrap things up, here are some terms you should know like the back of your gaming controller:

• Code Tracing : The detective work of following each line of code to see what variables are doing.
• Compound Assignment Operators : Shorthand operators that make coding life easier by combining arithmetic operations with assignment ( += , -= , *= , /= , %= ).
• Decrementing : The act of decreasing a value, aka -1ing something.
• Post-increment : The ++ operator that increases after using the current value.
• Pre-increment : The ++ operator that increases before using the new value.

Congratulations, you've made it through compound assignment operators! Now you're equipped with the knowledge to make your code more efficient and cleaner. Remember, practice makes perfect, and don't forget to use code tracing to keep track of those sneaky variables. 😎

Go on, conquer your coding challenges and ace that AP Computer Science A exam! Happy coding! 🌟

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Using multiple compound assignments in a single expression

I am preparing for a Java exam and I am trying to understand operator precedence and compound assignment operators in depth. I played around with a few expressions which use compound assignment during the evaluation of an expression. However, I do not fully understand why these give the result that they do.

What I expected:

Instead, a = 54 is the answer. So during all three evaluations of "a *= 2", the old value of a is used instead, so the final answer is a = 3 * 3 * 3 * 2 = 54.

However, this expression behaves differently:

If here the old value of c was used we would end up with c = 101 and d = 108. Instead, we end up with c = 2 * 5 + 100 = 110; meaning that in each of those evaluations of "c *= ..", the new (updated) value of c was used each time.

Question: why in the first example the original value of a gets used in the compound assignments, but in the second example the updated value of c gets used?

• operator-precedence
• compound-assignment

• By the way, first example doesn't need braces, assignments are evaluated from right to left –  Vasily Liaskovsky Commented Jul 19, 2023 at 12:30
• 1 This might be good fodder for an exam, and although these side effects can be handy at times they should be used judiciously as they can hide bugs that are hard to find. –  WJS Commented Jul 19, 2023 at 12:39
• I completely agree, @WJS –  Rauni Lillemets Commented Jul 20, 2023 at 12:46

3 Answers 3

According to §15.26.2. Compound Assignment Operators :

At run time, the expression is evaluated in one of two ways. [...] the value of the left-hand operand is saved and then the right-hand operand is evaluated

This explains the behavior for the first expression.

According to §15.18. Additive Operators :

The additive operators have the same precedence and are syntactically left-associative (they group left-to-right).

And §15.7. Evaluation Order :

The Java programming language guarantees that the operands of operators appear to be evaluated in a specific evaluation order, namely, from left to right.

Thus, the second expression is evaluated from left to right and side effects of updating the variables occur in that order.

"... why in the first example the original value of a gets used in the compound assignments, but in the second example the updated value of c gets used?"

For the first example, it's because it's already trying to assign a , so any inner assignments would not apply.

"... So during all three evaluations of "a *= 2", the old value of a is used instead ..."

Correct.  You are expecting it to do something like this,

For the second example, it is able to assign c since they are not part of one assignment.

a doesn't get update early in first example because a += x is equivalent to a = a + x and a as first operand of binary + is evaluated before second operand thus keeping initial value. And this condition preserves through the chain.

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Java Tutorial

Control statements, java object class, java inheritance, java polymorphism, java abstraction, java encapsulation, java oops misc.

In , assignment is used to assign values to a variable. In this section, we will discuss the . The is the combination of more than one operator. It includes an assignment operator and arithmetic operator or bitwise operator. The specified operation is performed between the right operand and the left operand and the resultant assigned to the left operand. Generally, these operators are used to assign results in shorter syntax forms. In short, the compound assignment operator can be used in place of an assignment operator. For example: Let's write the above statements using the compound assignment operator. Using both assignment operators generates the same result. Java supports the following assignment operators: Catagories Operator Description Example Equivalent Expression It assigns the result of the addition. count += 1 count = count + 1 It assigns the result of the subtraction. count -= 2 count = count - 2 It assigns the result of the multiplication. price *= quantity price = price * quantity It assigns the result of the division. average /= number_of_terms average = number_of_terms It assigns the result of the remainder of the division. s %= 1000 s = s % 1000 It assigns the result of the signed left bit shift. res <<= num res = res << num It assigns the result of the signed right bit shift. y >>= 1 y = y >> 1 It assigns the result of the logical AND. x &= 2 x = x & 2 It assigns the result of the logical XOR. a ^= b a = a ^ b It assigns the result of the logical OR. flag |= true flag = flag | true It assigns the result of the unsigned right bit shift. p >>>= 4 p = p >>> 4 Using Compound Assignment Operator in a Java Program CompoundAssignmentOperator.java Send your Feedback to [email protected] Help Others, Please Share Learn Latest Tutorials Transact-SQL Reinforcement Learning R Programming React Native Python Design Patterns Python Pillow Python Turtle Preparation Verbal Ability Interview Questions Company Questions Trending Technologies Artificial Intelligence Cloud Computing Data Science Machine Learning B.Tech / MCA Data Structures Operating System Computer Network Compiler Design Computer Organization Discrete Mathematics Ethical Hacking Computer Graphics Software Engineering Web Technology Cyber Security C Programming Control System Data Mining Data Warehouse Trending Now Foundational Courses Data Science Practice Problem Machine Learning System Design DevOps Tutorial Assignment Operators in Programming Assignment operators in programming are symbols used to assign values to variables. They offer shorthand notations for performing arithmetic operations and updating variable values in a single step. These operators are fundamental in most programming languages and help streamline code while improving readability. Table of Content What are Assignment Operators? Types of Assignment Operators Assignment Operators in C Assignment Operators in C++ Assignment Operators in Java Assignment Operators in Python Assignment Operators in C# Assignment Operators in JavaScript Application of Assignment Operators Assignment operators are used in programming to  assign values  to variables. We use an assignment operator to store and update data within a program. They enable programmers to store data in variables and manipulate that data. The most common assignment operator is the equals sign ( = ), which assigns the value on the right side of the operator to the variable on the left side. Types of Assignment Operators: Simple Assignment Operator ( = ) Addition Assignment Operator ( += ) Subtraction Assignment Operator ( -= ) Multiplication Assignment Operator ( *= ) Division Assignment Operator ( /= ) Modulus Assignment Operator ( %= ) Below is a table summarizing common assignment operators along with their symbols, description, and examples: Operator Description Examples = (Assignment) Assigns the value on the right to the variable on the left.  assigns the value 10 to the variable x. += (Addition Assignment) Adds the value on the right to the current value of the variable on the left and assigns the result to the variable.  is equivalent to  -= (Subtraction Assignment) Subtracts the value on the right from the current value of the variable on the left and assigns the result to the variable.  is equivalent to  *= (Multiplication Assignment) Multiplies the current value of the variable on the left by the value on the right and assigns the result to the variable.  is equivalent to  /= (Division Assignment) Divides the current value of the variable on the left by the value on the right and assigns the result to the variable.  is equivalent to  %= (Modulo Assignment) Calculates the modulo of the current value of the variable on the left and the value on the right, then assigns the result to the variable.  is equivalent to  Assignment Operators in C: Here are the implementation of Assignment Operator in C language: Assignment Operators in C++: Here are the implementation of Assignment Operator in C++ language: Assignment Operators in Java: Here are the implementation of Assignment Operator in java language: Assignment Operators in Python: Here are the implementation of Assignment Operator in python language: Assignment Operators in C#: Here are the implementation of Assignment Operator in C# language: Assignment Operators in Javascript: Here are the implementation of Assignment Operator in javascript language: Application of Assignment Operators: Variable Initialization : Setting initial values to variables during declaration. Mathematical Operations : Combining arithmetic operations with assignment to update variable values. Loop Control : Updating loop variables to control loop iterations. Conditional Statements : Assigning different values based on conditions in conditional statements. Function Return Values : Storing the return values of functions in variables. Data Manipulation : Assigning values received from user input or retrieved from databases to variables. Conclusion: In conclusion, assignment operators in programming are essential tools for assigning values to variables and performing operations in a concise and efficient manner. They allow programmers to manipulate data and control the flow of their programs effectively. Understanding and using assignment operators correctly is fundamental to writing clear, efficient, and maintainable code in various programming languages. Please Login to comment... Similar reads. Programming How to Get a Free SSL Certificate Best SSL Certificates Provider in India Elon Musk's xAI releases Grok-2 AI assistant What is OpenAI SearchGPT? How it works and How to Get it? Content Improvement League 2024: From Good To A Great Article Improve your Coding Skills with Practice What kind of Experience do you want to share? 277 IMAGES 025 Compound assignment operators (Welcome to the course C programming) Try an Order of Operations Maze! The Compound Assignment Operators PPT Compound Assignment Operators PPT COMMENTS 1.5. Compound Assignment Operators After doing this challenge, play the Operators Maze game. See if you and your partner can get the highest score! 1.5.2. Summary¶ Compound assignment operators (+=, -=, *=, /=, %=) can be used in place of the assignment operator. The increment operator (++) and decrement operator (--) are used to add 1 or subtract 1 from the stored value of a ... 1.5. Compound Assignment Operators 1.5. Compound Assignment Operators ¶. Compound assignment operators are shortcuts that do a math operation and assignment in one step. For example, x += 1 adds 1 to x and assigns the sum to x. It is the same as x = x + 1. This pattern is possible with any operator put in front of the = sign, as seen below. + shortcuts. - shortcuts. Compound assignment operators in Java The compound assignment operators are +=, -=, *=, /=, %= etc. The. 2 min read. Java Assignment Operators with Examples. Operators constitute the basic building block of any programming language. Java too provides many types of operators which can be used according to the need to perform various calculations and functions, be it logical ... Compound Assignment Operators integerOne *= 2; The "* = 2" is an example of a compound assignment operator, which multiplies the current value of integerOne by 2 and sets that as the new value of integerOne. Other arithmetic operators also have. compound assignment operators. as well, with addition, subtraction, division, and modulo having +=, -=, /=, and %=, respectively. 1.5 Compound Assignment Operators #APCSA #CSAwesome #LearnJavaCSAwesome (course.csawesome.org) is a College Board AP Computer Science A curriculum to teach Java programming. Compound assignme... 1.5. Compound Assignment Operators 1.5. Compound Assignment Operators¶. Compound assignment operators are shortcuts that do a math operation and assignment in one step. For example, x += 1 adds 1 to the current value of x and assigns the result back to x.It is the same as x = x + 1.This pattern is possible with any operator put in front of the = sign, as seen below. PDF Compound assignment operators The compound operators are different in two ways, which we see by looking more precisely at their definition. The Java language specification says that: The compound assignment E1 op= E2 is equivalent to [i.e. is syntactic sugar for] E1 = (T) ((E1) op (E2)) where T is the type of E1, except that E1 is evaluated only once. E1 is evaluated once. CS106B Mazes A maze is a twisty and convoluted arrangement of paths that challenge the solver to find a route from the entry to the exit. This assignment is about using ADTs to represent, process, and solve mazes. ... The assignment operator work as expected for all our ADTs. Assigning from one Stack/Vector/Set/etc to another will create a copy of the ADT ... Compound Assignment Operators Compound assignment operators are shorthand notations that combine an arithmetic operation with the assignment operator. They allow you to perform an operation and assign the result to a variable in a single step. All Subjects. Light. AP Computer Science A. Unit 1 - Primitive Types. Unit 2 - Using Objects ... Java Compound Operators Compound Assignment Operators. An assignment operator is a binary operator that assigns the result of the right-hand side to the variable on the left-hand side. The simplest is the "=" assignment operator: int x = 5; This statement declares a new variable x, assigns x the value of 5 and returns 5. Compound Assignment Operators are a shorter ... Compound Assignment Operators Flashcards Compound Assignment Operator. Shortcuts that do a math operation and assignment in one step. 1 / 4. 1 / 4. Flashcards; Learn; Test; Match; Q-Chat; Created by. kestephenson56 Teacher. Share. Share. Get better grades with Learn. 82% of students achieve A's after using Learn. Study with Learn. Students also viewed. Expected Value. 13 terms. Compound Assignment Operators Compound assignment operators are shortcuts that do a math operation and assignment in one step. For example, x += 1 adds 1 to x and assigns the sum to x. It is the same as x = x + 1. This pattern is possible with any operator put in front of the = sign, as seen below. The most common shortcut operator ++, the plus-plus or increment operator ... Compound Assignment Operators Compound Assignment Operators: Shorthand operators that make coding life easier by combining arithmetic operations with assignment ( +=, -=, *=, /=, %= ). Decrementing: The act of decreasing a value, aka -1ing something. Post-increment: The ++ operator that increases after using the current value. Using multiple compound assignments in a single expression Additive Operators: The additive operators have the same precedence and are syntactically left-associative (they group left-to-right). And §15.7. Evaluation Order: The Java programming language guarantees that the operands of operators appear to be evaluated in a specific evaluation order, namely, from left to right. #21 Study with Quizlet and memorize flashcards containing terms like What do compound assignment operators do?, What are the compound assignment operators for the following: Addition Subtraction Multiplication Division Modulo, SECTION REVIEW1A: #1) You are also in charge of keeping track of how many cookies there are at the bake sale. This value is represented by the number of numCookies. Compound Assignment Operator in Java The compound assignment operator is the combination of more than one operator. It includes an assignment operator and arithmetic operator or bitwise operator. The specified operation is performed between the right operand and the left operand and the resultant assigned to the left operand. Generally, these operators are used to assign results ... Assignment Operators in Programming Assignment operators are used in programming to assign values to variables. We use an assignment operator to store and update data within a program. They enable programmers to store data in variables and manipulate that data. The most common assignment operator is the equals sign (=), which assigns the value on the right side of the operator to ... 1.5 Compound Assignment Operators.pdf Document 1.5 Compound Assignment Operators.pdf, Subject Computer Science, from Kang Chiao International School, Length: 1 pages, Preview: 1.5.1 Code Tracing Challenge and Operators Maze x += 1 is the same as x = Please share free course specific Documents, Notes, Summaries and more! essay homework paper phd review speech study thesis