C program different Output on different Compilers?

Question

Here is the C program which is giving different output depending on Compiler used:

#include<stdio.h>
int main()
{
    int i = 5,j;
    j = ++i + i++ + ++i + i++;
    printf("%d",j);
    return 0;
}

Check the output on the following link.

http://imgur.com/z9aMSwj,Vwx3P9S

http://imgur.com/z9aMSwj,Vwx3P9S#1

My question is what is the technical reason that output is different?

http://stackoverflow.com/questions/3812850/output-of-multiple-post-and-pre-increments-in-one-statement — AntonH, Mar 10 '14 at 17:52
Why would you expect the output to be consistent? Do you think all compilers behave exactly the same? Why are there different compilers then? — abelenky, Mar 10 '14 at 17:58

score 0 · Answer 1 · answered Mar 10 '14 at 17:52

0

The technical reason is that there is no defined behaviour for such an operation, allowing compilers to handle this kind of an order as they wish. Such cases are generally referred as Undefined Behaviour.

answered Mar 10 '14 at 17:52

Utkan Gezer

3,009
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score 0 · Answer 2 · answered Mar 10 '14 at 17:52

0

According to C language, expressions like ++i + i++ + ++i + i++ have undefined behavior.

answered Mar 10 '14 at 17:52

Arjun Sreedharan

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There are a lot of questions asked in placement papers of this type, so can you please tell me what should be my approach to deal with these questions? – Akash Rana Mar 10 '14 at 17:57
@user3402873 get familiar with the concept of sequence points http://en.wikipedia.org/wiki/Sequence_point – Arjun Sreedharan Mar 10 '14 at 18:01

score 0 · Answer 3 · answered Mar 10 '14 at 17:59

In a i++ expression, two things happen. First, the expression is evaluated to i, then i is incremented later. The behaviour of 'later' is undefined. If there are 4 such expressions in a statement, there are countless possible orders to evaluate the increments and decrements.

C program different Output on different Compilers?

3 Answers3