Post/pre increments in 'printf' [duplicate]

This question already has answers here: Closed 12 years ago.

Possible Duplicates:

Output of multiple post and pre increments in one statement

Post-increment and pre-increment in 'for' loop

The following code snippet

int i=0;
printf("%d %d",i++,i++);

gives the output

1 0开发者_开发问答

I can understand that, but the following

int i=0;
printf("%d %d",++i,++i);

gives the output

2 2

Can someone explain me the second behavior?

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Both printfs invoke undefined-behavior. See this : Undefined behavior and sequence points

Quoted from this link:

In short, undefined behaviour means anything can happen from daemons flying out of your nose to your girlfriend getting pregnant.

For newbies : Don't ever try to modify values of your variables twice or more in a function call argument-list. For details, click here to know what it means. :-)

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They're both undefined behaviour. Modifying the variable i more than once is undefined. Also, C++ or C? You need to make up your mind as the behaviour of pre-increment I believe is different between them.

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You got what called 'undefined behaviour', because you are changing the same variable more than once between sequence points. Another compiler can give you different results.