Java Puzzle #1: Pre- and Postincrement

The puzzle

What is the output of the following piece of code?

public class test {
    public static void main(String[] args) {
        int i = 1;
        i += ++i + i++ + ++i;

        int j = 1;
        j += ++j + j++ + ++j;

        int k = 1;
        k += k++ + k++ + ++k;

        int m = 1;

        System.out.println("i = " + i);
        System.out.println("j = " + j);
        System.out.println("k = " + k);
        System.out.println("m = " + (m += 1));
    }
}

. . . . . . . . . . . . . . . . . . . . . .

Answer

The output is:

i = 9
j = 9
k = 8
m = 2

Explanation

Part one

First, take a look at statements of this structure:

i += s

where i is the integer and s is a statement (e.g. ++i). This gets evaluated to

i = a + s

Source: docs.oracle.com

Part two

Lets take a look at pre- and postincrement in Java.

You can quite easily figure out what the different increments do by this snippet:

public class test {
    public static void main(String[] args) {
        int i = 0;
        int j = 0;
        int k = 0;

        System.out.println("i = " + ++i);
        System.out.println("i = " + i);

        System.out.println("j = " + j++);
        System.out.println("j = " + j);

        System.out.println("k = " + (k += 1));
    }
}

Output:

i = 1
i = 1
j = 0
j = 1
k = 1

Line 7 adds +1 to i and returns the value. Line 10 returns the value of j and adds +1 to j. Line 13 adds +1 to k and returns k.

Lets return to the original puzzle. Java parses your code from left to right (Source 1, Source 2).

Most important:

Evaluation of an expression can also produce side effects, because expressions may contain embedded assignments, increment operators, decrement operators, and method invocations.

So:

int i = 1;
i += ++i + i++ + ++i;

is the same as

i = ((i + (++i)) + (i++)) + (++i);

The first ++i increments i to 2 and returns 2. So you have:

i = 2;
i = ((1 + 2) + (i++)) + (++i);

The i++ returns 2, as it is the new value of i, and increments i to 3:

i = 3;
i = ((1 + 2) + 2) + ++i;

The second ++i increments i to 4 and returns 4:

i = 4;
i += ((1 + 2) + 2) + 4;

So you end up with 9.

A parse tree of this evaluation would look like this:

The explanation for the other three ones is similar.

See also

• Increment and decrement operators
• Why avoid increment (“++”) and decrement (“--”) operators in JavaScript?
• Pre- and postincrement in Java