CODEWITHSUNDEEP
c
cdummy
cdummy

Reputation: 455

increment operator within conditional operator in c

I expected b to be 3.
is this an undefined behavior? or the result could be predicted?
but i thought b would be 3.
and c would be 2. the output i got is 1 2 1
please explain 

#include<stdio.h>
int main()
{
int a = 0;
int b = 1;
int c = (a++ >b++ )? a++ : b++;
printf("%d %d %d",a,b,c);
return 0;
}

Upvotes: 1

Views: 739

Answers (2)

Matthew Flaschen
Matthew Flaschen

Reputation: 284927

That's what I get, and I agree it's guaranteed to be that. There is a sequence point after evaluating the first operand (the condition) before going to the second or third operand. So it goes like this:

(a++ >b++ )

evaluates to:

0 > 1

which is 0.

After that, a is 1 and b is 2.

Since it was false:

b++

is evaluated. The result is 2 (which is assigned to c), and afterwards b is 3.

If that's the exact code, your compiler is buggy. It's not even a question of order. Even if the third operand were evaluated before the first (which would be wrong), b should still be 3.

I am using GCC 4.6.3, but the result will be the same in all standards-compliant compilers.

Upvotes: 4

cnicutar
cnicutar

Reputation: 182674

It's defined behavior, there's a sequence point between the first operand of ?: and second or third one.

So after evaluating a++ < b++, a = 1 and b = 2. Then the third operand gets selected. Thus c gets assigned b++. So c = 2 and then b = 3.

The C11 standard says:

6.5.15 Conditional operator

The first operand is evaluated; there is a sequence point between its evaluation and the evaluation of the second or third operand (whichever is evaluated).

Upvotes: 2

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