int main()
{
int b=0,a=1;//initialize a and b
b=++a + ++a;// calculate assign the value of b
print f("%d",b);
return 0;
}
- what is the value of b?
- and what is the calculation for it?
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Arijit
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1This question and the famous *== vs equals in Java* are the most two asked question on earth. – Maroun Aug 06 '13 at 11:04
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1`undefined behaviour` for the win – user2485710 Aug 06 '13 at 11:05
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I wonder why question papers have these type of questions ? – P0W Aug 06 '13 at 11:12
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Maybe SO should automagically prevent posting of questions containing `++a + ++a` and such. – devnull Aug 06 '13 at 11:14
2 Answers
2
This is Undefined Behaviour.Lack of sequence point.
For more info have a look here and output-of-multiple-post-and-pre-increments-in-one-statement.
0
It would seem that having a sequence point is immaterial in the expression b=++a + ++a;
That is, whether the first ++a is evaluated first or the second ++a is evaluated first in either case a is incremented twice and then the + operator takes effect, so the eventual equation is either b = 2 + 3; or b = 3 + 2 thus b = 5.
When I get home I will try this on with my C compiler.
Blastfurnace's comment about both being evaluated before the + operator takes effect is correct, and now that I think about it, obvious.
That is, + is lower in precedence than ++a. It could be argued that this statement is NOT ambiguous in that switching the order of evaluation (R to L or L to R, both with respect to precedence) results in the same answer.
No one will claim this is well written code, interesting on several points for a discussion, but not something that should be endorsed.
JackCColeman
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...or both side are evaluated before the addition and the result is `3 + 3`. Either way this code is awful and there is no point trying to reason about it. – Blastfurnace Aug 07 '13 at 00:25
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in my c compiler the value of b is showing is 6. here a single memory named "a" is shared for both times. for first ++a the vale of a = 2 and for 2nd ++a ,a=3 then b=3+3=6. – Arijit Sep 29 '13 at 14:22
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