问题
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
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?
回答1:
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. :-)
回答2:
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.
回答3:
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.
来源：https://stackoverflow.com/questions/4706404/post-pre-increments-in-printf