`
a = 10;
int *ptr = &a;
printf("%d %d\n", a, ++*ptr);
`
The output is - 11 11
How is it evaluated??
Asked
Active
Viewed 58 times
0
Ayush Chothe
- 13
- 1
-
1@EugeneSh. there's no sequence point between them, so UB – Paul Hankin Dec 05 '22 at 19:01
-
@PaulHankin OK, right. Was under wrong impression that evaluation of each parameter is sequenced as a full expression – Eugene Sh. Dec 05 '22 at 19:02
-
1The order of evaluation is not specified. That is the problem. Imo sequence point is there. – 0___________ Dec 05 '22 at 19:15
-
@0___________ The sequence point is when entering the function, but there isn't between evaluation of the two parameters – Eugene Sh. Dec 05 '22 at 21:18
1 Answers
0
This is undefined behaviour, so the result could be anything. There is no sequence point in the line, which means that both operations are unsequenced - either argument could be evaluated first, or both simultaneously (see https://en.wikipedia.org/wiki/Sequence_point and John Bollinger's comment).
For example, when evaluating with the clang compiler, this is the output:
<source>:5:26: warning: unsequenced modification and access to 'a' [-Wunsequenced]
printf("%d %d\n", a, ++a);
~ ^
1 warning generated.
Execution build compiler returned: 0
Program returned: 0
2 3
See this answer for more: Order of operations for pre-increment and post-increment in a function argument?
CoderMuffin
- 519
- 1
- 10
- 21
-
2The "For example, if `a` is evaluated first" is nonsensical and obscures the point. The inherent issue is that `a` is *not* evaluated before `++*ptr`, nor after, nor even at the same time. The evaluations of those two expressions are *unsequenced* relative to each other as far as C semantics are concerned. And because they are unsequenced and one has a side effect on an object read by the other, the behavior is undefined. – John Bollinger Dec 05 '22 at 19:24
-
Thanks for the info, I have removed it and tried to improve the answer – CoderMuffin Dec 06 '22 at 07:43