Incrementation operator functionality. How does it work?

Question

Can someone tell me why this code results in 1?

What I think should happen is myInt gets modded by 10, resulting in 1, then myInt gets incremented and should become 2. However, it seems the incrementation is discarded.

int myInt = 21;

myInt = myInt++ % 10;

System.out.println( "myInt: " + myInt );

I'm not sure what sort of answer you're looking for. Your intuition was wrong, and actually the increment happens first, then the assignment. You just verified this. — Ismail Badawi, Dec 02 '13 at 05:32

score 2 · Answer 1 · answered Dec 02 '13 at 05:37

2

Google for difference between postincrement and preincrement.

This will work for your case

int myInt = 21;

myInt = ++myInt % 10;

System.out.println( "myInt: " + myInt );

answered Dec 02 '13 at 05:37

Suganthan Madhavan Pillai

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score 0 · Answer 2 · answered Dec 02 '13 at 05:35

0

Rather than theoretically, I think it would be good to explain through example.

there are two type of increment

a++ (post increment)

++a (pre increment)

If a = 10;

i=++a + ++a + a++; =>
i=11 + 12 + 12; (i=35)

i=a++ + ++a + ++a; =>
i=10 + 11 + 12; (i=33)

answered Dec 02 '13 at 05:35

vinay kumar

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At least in C, `++a + ++a + a++` would be undefined behavior. How is this in Java? – Felix Kling Dec 02 '13 at 05:36
@FelixKling It's not undefined in Java. (http://docs.oracle.com/javase/specs/jls/se7/html/jls-15.html#jls-15.7) – Ismail Badawi Dec 02 '13 at 05:40
@FelixKling - it works in java. – vinay kumar Dec 02 '13 at 05:41
@Ismail: This doesn't give a definite answer about those operators though (at least I don't see it). But after browsing some questions on SO, it seems the behavior is indeed defined. – Felix Kling Dec 02 '13 at 05:47
@vinaykumar: It "works" in C as well, but the actual result can differ from machine to machine. – Felix Kling Dec 02 '13 at 05:48
@FelixKling - I don't have indepth knowledge on C, but very curious to know how it differs from machine to machine, whether it will throw an exception or, will give a wrong result? – vinay kumar Dec 02 '13 at 05:58
Have a look at: https://en.wikipedia.org/wiki/Undefined_behavior. It doesn't throw an exception, you would just get different results. – Felix Kling Dec 02 '13 at 06:04
@FelixKling Thank you. – vinay kumar Dec 02 '13 at 06:07

score 0 · Answer 3 · answered Dec 02 '13 at 05:39

I believe the operator "variable++" is a post-increment operator. As a result the original value is returned BEFORE incrementing.

So in your case:

myInt++ returns 21
myInt++ is process, myInt now equals 22
21 % 10 is processed and assigned to myInt, myInt equals 1

Incrementation operator functionality. How does it work?

3 Answers3