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javaoperators
Toraen
Toraen

Reputation: 3

Java Postfix Increment Operator Behavior

In the sample code below, why is the end value different after each method. I would expect both to display "i = 1"

public class Test {

    public int i = 0;    

    public static void main( String args[] ) {        
        Test t = new Test();
        t.test();
    }

    public void test() {

        i = 0;
        System.out.println( "[start a] i = " + i );
        doSomethingA( i++ );
        System.out.println( "[end   a] i = " + i );

        System.out.println( "---------------------" );

        i = 0;
        System.out.println( "[start b] i = " + i );
        doSomethingB( i++ );
        System.out.println( "[end   b] i = " + i );

    }

    // Direct assignment of passed value
    public void doSomethingA( int x ) {
        i = x;
    }

    // Equation of passed value
    public void doSomethingB( int x ) {
        i += x;
    }

}

The results are:

[start a] i = 0
[end   a] i = 0
---------------------
[start b] i = 0
[end   b] i = 1

Why does it matter what I do to 'i' in the methods, shouldn't it increment by 1 after the method ends?

In both cases I am assigning the value of 'i' to 0 inside the method.

Thanks

Upvotes: 0

Views: 105

Answers (2)

Pshemo
Pshemo

Reputation: 124235

Add 

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

at start of doSomethingA and doSomethingB methods and you will see that in both cases 

i=1, x=0

This is because in 

method(i++);

first current value of i is passed to method making x = 0, then field i is incremented (to 1) and then body of method is executed. So when inside methods body you do

i = x;

you are setting i to 0

but if you do 

i += x;

you are adding 0 to current value of i which is 1 which doesn't change value of i.

Upvotes: 0

rgettman
rgettman

Reputation: 178263

Here's what happens when calling both methods from test.

First.

i starts out at 0 and 0 is printed. i++ is evaluated. Because it's post-increment, the expression value is the old value, 0, so that is what's passed to doSomethingA. The post-increment leaves i at 1. doSomethingA assigns x (0) back to i, so 0 is printed.

Second.

i starts out at 0 and 0 is printed. i++ is evaluated. Because it's post-increment, the expression value is the old value, 0, so that is what's passed to doSomethingB. The post-increment leaves i at 1. doSomethingB adds x (0) to i, so i remains 1, and 1 is printed.

Upvotes: 1

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