#include <stdio.h>
#include <stdlib.h>
int main()
{ int i;
i=0;
printf("%d,%d,%d",i++,++i,i++);
return 0;
}
Output: 2,3,0 I expect the output to be:2,2,0 Can someone explain how it actually gets evaluated
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Manisha P
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Please elaborate your expectation. Why are you expecting "2,3,0"? I admit I expected "0,2,2". "0" because it is a postincrement, not done before reading the value. First "2" because the previous postincrement and a preincrement is effective when reading. Second "2" because, again, the postincrement did not get effective before reading. – Yunnosch Apr 01 '17 at 20:46
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Why are you expecting "3,2,0"? Obviously is what I meant to ask. – Yunnosch Apr 01 '17 at 20:54
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Possible duplicate of (http://stackoverflow.com/questions/376278/parameter-evaluation-order-before-a-function-calling-in-c) – Yunnosch Apr 01 '17 at 20:57
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sorry, expected is 2,2,0. But what i get after compilation is 2,3,0.. and i don't know how is that – Manisha P Apr 02 '17 at 19:51
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The question has been classified as "duplicate", i.e. the question and answer is already on Stackoverflow. The official other question discusses undefined behaviour. My recommendation is on why/that evaluation of parameters to functions is unpredictable (from compiler to compiler); quite interesting. Basically compilers can do anyting they like for optimisation. You must not attempt anything which requires prediction of parameter evaluation order. – Yunnosch Apr 02 '17 at 20:43