Why is “i” variable not getting incremented twice in my program?

Question

Why is "i" variable getting incremented twice in my program?

I modified the question but why is the output different. shouldn't it be same?.?

Code :-

#include<stdio.h>
#define MAX(x,y) (x)>(y)?(x):(y)
void main(void)
{
    int i = 10;
    int j = 5;
    int k = 0;
    k == MAX(i++, ++j);
    printf("%d %d %d",i,j,k);
}

Output :- 11 7 0

Shouldnt the output be 12 6 0

I think this is undefined behaviour, see e.g. http://stackoverflow.com/questions/4176328/undefined-behavior-and-sequence-points — Erich Kitzmueller, Aug 08 '14 at 10:53
@ammoQ It is not. The conditional operator has a sequence point between the condition and the evauluation of the 2nd or 3rd operand — Lundin, Aug 08 '14 at 10:54
Even if the macro didn't do weird things, `k == MAX(i++, ++j);` doesn't do anything useful. — Retired Ninja, Aug 08 '14 at 10:58
Are you sure you have the `>` operator in the macro, and not the `<`? — Some programmer dude, Aug 08 '14 at 10:58
+1 btw, because it is funny to see how many of our C gurus that failed to immediately point out why the results turned out as they did, yours sincerely included :) Once you properly expand the whole line of code and not just the macro, it all becomes clear. — Lundin, Aug 08 '14 at 11:13

mafso · Answer 1 · 2014-08-16T22:35:48.630

Use braces around your macro definitions:

#define MAX(x,y) ( (x)>(y)?(x):(y) )

The expanded expression (with the original macro definition)

k == (i++)>(++j)?(i++):(++j);

groups as

(k == i++ > ++j) ? i++ : ++j;

i++ > ++j evaluates to 1, which is unequal to k (k is 0), so the third operand of ?: (++j) is evaluated. So, i is incremented once, j twice, and k wasn’t changed since its initialization, hence your output.

Lundin · Accepted Answer · 2014-08-08T11:18:54.963

The problem is that the macro expands into this:

k == i++ > ++j ? i++ : ++j;

Operator precedence dictates that > has higher prio than == which has higher prio than ?: So the above is equivalent to this:

(k == (i++ > ++j)) ? i++ : ++j;

i++ and ++j are executed first. Since i++ > j++ gets evaluated to true (1), we get this:

(k == 1) ? i++ : ++j;

Then since k is 0, k == 1 is evaluated to false (0). We get:

0 ? i++ : ++j;

And therefore ++j is executed once more.

(As a side note, this expression is well-defined, because the conditional operator has a sequence point between the condition and the evauluation of the 2nd or 3rd operand.)

score 2 · Answer 3 · answered Aug 08 '14 at 11:05

k == MAX(i++, ++j);

is replaced as

k == (i++)>(++j)?(i++):(++j);

Here i=10,j=5. so

k == (10++)>(6)?(11++):(++6);

so it that expression while checking condition i is incremented once after checking the condition due to post increment and j also incremented once. But 10 > 6 condition is true.

(k == 1 )?(i++):(++j); // here i=11 and j=6

(k == 1 )? (11++):(++6);

so here k==1 condition fails. it will not evaluate i++ in that expression, it will evaluate ++j and return it. here i is remains 11 and j become 7.

So you will get 11 7 0 as output.

Why is “i” variable not getting incremented twice in my program?

3 Answers3