I couldn't figure out the reason for the output I got:-
int ar[5] = {1, 3, 5, 7, 9};
int *p = ar;
printf("%d\t%d\n", *p, *(p++));
output: 3 1
but i expected : 1 3 as p points to the 1st element and p++ points to the 2nd element.
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user3248186
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Hint: `printf("%d\n", *p);` prints 1, as you expected. – Maroun Jan 29 '14 at 10:14
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2I do not think this is UB. I believe it is *implementation specific*, which is different. (Not any better to rely on, though) – user694733 Jan 29 '14 at 10:16
3 Answers
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Order of evaluation for function arguments is unspecified.
pmr
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@BlueMoon Paragraph 5/4 `Between the previous and next sequence point a scalar object shall have its stored value modified at most once by the evaluation of an expression.` I only see one modification, but I might be wrong here. – pmr Jan 29 '14 at 10:18
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order of evaluation from right to left evaluate *(p++) first and increase the address then evaluate *p which now point to next element.Try *(p+1) instead of it.
Jayesh Bhoi
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i know that *(p+1) gives 3, but i wanna know what's wrong with my code – user3248186 Jan 29 '14 at 10:38
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nothing wrong in your code! It just parameter evaluation result and note that parameter evalution is undefined. For looking more comment on this topic go
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This gives a nice example how order of evaluation for function arguments is not granted.
