Square using define directive not working

Question

I implemented a squaring define directive as this :

#include <stdio.h>

#define SQR(y) ((y)*(y))

int main() {
    int a;
    printf("Enter : ");
    scanf("%d",&a);
    a = SQR(a);
    printf("The square is: %d\n",SQR(a));
    return 0;
}

But when I execute it, I don't get any error, but it is giving wrong answer every time.

Why is it giving 4th power instead of second power of input?

"*But when i execute it, I dont get any error*" So why do you have the tags [tag:compiler-errors] and [tag:runtime-error]? — Spikatrix, May 10 '15 at 07:37

score 2 · Accepted Answer · answered May 10 '15 at 07:36

You are squaring it twice.

In your code:

a = SQR(a);
printf("The square is: %d\n",SQR(a));

this first makes a = a*a
Now you print SQR((a*a)*(a*a)), or a*a*a*a

You can replace SQR(a) in printf with a, or remove the a = SQR(a)

Try this :

#include <stdio.h>
#define SQR(y) ((y)*(y))

int main() {
    int a;
    printf("Enter : ");
    scanf("%d",&a);
    printf("The square is: %d\n",SQR(a));
    return 0;
}

score 0 · Answer 2 · answered May 10 '15 at 07:36

0

Because you are writig ((a)*(a)) twice in your code.

a = SQR(a); //<-- Here a = a*a printf("The square is: %d\n",SQR(a)); //<-- and here print a*a

answered May 10 '15 at 07:36

Vagish

2,520
19
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score 0 · Answer 3 · answered May 10 '15 at 07:36

0

You calculate the square two times. You first square a here:

a = SQR(a);

and then, square this number again in the printf:

printf("The square is: %d\n",SQR(a));

Fix the problem by replacing the above printf with:

printf("The square is: %d\n", a);

answered May 10 '15 at 07:36

Spikatrix

20,225
7
37
83

@weston I get red dialog box which reads : "Requires 15 reputation" – Doomsday 666 May 12 '15 at 17:06
@Doomsday666 sorry, I didn't know/remember there was a limit – weston May 12 '15 at 19:35

score 0 · Answer 4 · answered May 10 '15 at 07:51

you are misunderstanding the usage of c macro.

There is no issue with your macro definition. The issue is with the usage.

a = SQR(a);

In above line, you calculate the square of a and assign result to a again. So after above line execute the value of a is square of input value.

Then again in below line, you are calling macro on a.

printf("The square is: %d\n",SQR(a));

Then you prints the square of the current value of a. That value is not wt you need to get.

For example, if you input 2, you will get 16, but you need 4 as the answer.

Square using define directive not working

4 Answers4