int main()
{
int a=10;
if (a == a--)
printf("true1\t\n");
a=10;
if(a == --a)
{
printf("true2");
}
return 0;}
as in the second if condition a=10 and --a is 9 i.e 10 != 9 so how come the second condition is evaluated ?
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James
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Google for **Undefined Behaviour**; I'm sure this is a duplicate. – pmg Oct 05 '14 at 17:31
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3http://stackoverflow.com/questions/4176328/undefined-behavior-and-sequence-points – phuclv Oct 05 '14 at 17:33
2 Answers
7
The value of --a is the previous value of a minus 1.
Furthermore that expression has the side-effect of changing the value of a.
The left part of the comparison is the value of a ... but is it the value of a before or after the side-effect has been applied?
The C Standard does not force the sequence of checking the value and applying the side-effect; and says that reading the value of a variable and changing its value without an intervening sequence point is Undefined Behaviour.
Basically there is a sequence point at every ; in the program (it's not as straightforward); your expression (a == --a) does not have sequence points.
pmg
- 106,608
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I believe the next rule is also valid in C as in Java : The --x has a higher priority than x--
a=10;
printf(a--) // should be showing 10;
a=10;
printf(--a) //should be showing 9
Stefanel S
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Though it is true (I didn't downvote), a) not really related to question, b) it's not valid C. – Grzegorz Szpetkowski Oct 05 '14 at 19:05