Unexpected output printf statement

Question

Why outputs of i and j in the following two printf()s are different?

#include <cstdio>
#define MAX(x,y) (x)>(y)?(x):(y)
int main()
{
    int i=10,j=5,k=0;
    k==MAX(i++,++j);
    printf("%d %d %d\n",i,j,k);
    i=10,j=5,k=0;
    k=MAX(i++,++j);
    printf("%d %d %d\n",i,j,k);
    return 0;
}

I think only k should differ, but it is showing different output for all, i, j and k, in printf() statement. What is the reason?

Output :

11 7 0
12 6 11

Did you mean sing `=` in line `k==MAX(i++,++j);`? – Mohit Jain Jun 16 '15 at 07:35 — Mohit Jain, Jun 16 '15 at 07:35
So what is your expected output for each print statement? – ilent2 Jun 16 '15 at 07:42 — ilent2, Jun 16 '15 at 07:42
@ilent2, question edited to make it clear – shreyans800755 Jun 16 '15 at 13:17 — shreyans800755, Jun 16 '15 at 13:17

Mohit Jain · Accepted Answer · 2015-06-16T07:45:26.440

k==MAX(i++,++j); is translated to:

  k==(i++) > (++j) ? (i++) : (++j);
/* 2       1       3  : Precedance */

In your case i = 10 and j = 5, so the comparision would be 10 v/s 6 which would be true or 1 with side effect i becomes 11 and j becomes 6. (Recall that: Macros are not functions)

As k = 0 and not equal to 1, k==(i++) > (++j) would be false and ++j would be executed. Causing j to be incremented to 7.

In latter case, there is no modification to the value of variables.

When can I expect same output in both case? Even if you change the macro with a function, your arguments have the side effects, which would cause the output 11 6 for i j, so you are changing the variables using ++ operator, don't expect both statements to have the same values.

score 1 · Answer 2 · edited May 23 '17 at 11:58

Let's analyse the output from first printf()
1. Assuming you meant = instead of == in your code,
```
k==MAX(i++,++j);
```
  gets translated to
```
 k = (i++) > (++j) ? (i++): (++j)
```
  which gets evaluated changing all the values of i, j and k.
2. Even if you meant ==, as mentioned in the answer by Mr. Mohit Jain, you'll face similar issue with precedence.
Have a look at operator precedence for better understanding.
Let's analyse the output from second printf()

In the second printf() statement, you're re-setting the values of i, j, k and not using the MACRO. So, those values will get printed. Simple.

So, TL;DR, before first printf(), the MACRO changes all the three values for i, j, k. So, your output for first and second printf() differs for all the three variable values.

FWIW, if you meant the MACRO to behave as a ternary conditional expression , you need to change your MACRO definition to

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

To avoid issues with operator precedence after MACRO expansion.

ouah · Answer 3 · 2015-06-16T07:39:29.957

1

You have side-effects.

MAX(i++,++j)

is expanded to

(i++)>(j++)?(i++):(j++)

And i is modified twice in the macro call.

Moreover, you have to add parens to your macro definition:

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

edited Jun 16 '15 at 07:39

answered Jun 16 '15 at 07:37

ouah

142,963
15
272
331

1

how? after `?` there __is__ a sequence point, isn't it? (Please correct me if i'm wrong) – Sourav Ghosh Jun 16 '15 at 07:38
@SouravGhosh correct, I edited my answer – ouah Jun 16 '15 at 07:39

score 1 · Answer 4 · answered Jun 16 '15 at 07:38

1

i++ and ++j both increment i and j, so you change their values before printing.

answered Jun 16 '15 at 07:38

Simon

9
3

score 0 · Answer 5 · answered Jun 16 '15 at 07:50

The macro is a red herring. OP did not ask about k but "Why output of i and j in following two printf are different?" The answer is because i and j have been modified between setting their values and printing, in the first case, but not in the second case.

Unexpected output printf statement

5 Answers5