Understanding the working of Unary Operators

Question

What will be the output of the following code? Online compilers give the result: 12 6 11. Can someone tell me how?

#include<stdio.h>

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

main()
{
    int i = 10, j = 5, k = 0;
    k = MAX(i++, ++j);
    printf("%d %d %d", i, j, k);
}

Wrong duplicate, @KerrekSB (though it most likely is a duplicate). No UB in here. — Kijewski, Jul 31 '14 at 08:22
@Kay: You're right, the evaluation is sequenced -- I'm sorry! — Kerrek SB, Jul 31 '14 at 08:27
Actually I don't think there's any undefined behavior in this code, the duplicate seems all wrong ... Perhaps writing your code like this: `k = (i++) > (++j) ? (i++) : (++j)` and get rid of the macro would make it clearer ... — dragosht, Jul 31 '14 at 08:27
@dragosht: Yes, my fault - there's no UB, since the evaluations are sequenced. — Kerrek SB, Jul 31 '14 at 08:28
@Kay: (Yes, I saw an identical question yesterday or two days ago...) — Kerrek SB, Jul 31 '14 at 08:29

Maroun · Accepted Answer · 2014-07-31T08:30:29.967

2

It gets translated to:

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

Now it's easy to tell why..

i++ makes i 11, ++j makes j 6. Now we have:

10 > 6

Which is true, then we return i++, which is 12 now.

Use the -E flag to see the output of the pre-processor.

edited Jul 31 '14 at 08:30

answered Jul 31 '14 at 08:22

Maroun

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Maybe you could rewrite this a little bit. `?:` → `if else`, `i++` → `++i; … i` – Kijewski Jul 31 '14 at 08:24

score 0 · Answer 2 · answered Jul 31 '14 at 08:56

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

This code will be replaced by

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

when checking the condition (i++)>(++j), i is post increment(after the operation it will increment) and j is pre-increment( increment the j and substitute it). so while checking it will take (10) > (6), it is true! after this check i will become 11.

k = (i++)>(++j)?(i++):(++j); --> k = (i++)>(++j)?((11)++):(++j); // note after condition check i becomes 11.
// due to post increment after assigning the value to k, i will be incremented to 12. but the last ++j wont excuted.

So you will get output 12 6 11.

else you can rewrite the k = (i++)>(++j)?(i++):(++j); as

if((i++)>(++j)) //when condition check i=10, j=6 after check i=11,j=6
k=i++; // k= 11++; after assigning 11 to k i will be incremented to 12
else
k=++j; // this wont be executed

Understanding the working of Unary Operators

2 Answers2