output of c code

Question

Why output is giving 3 , expecting -3. How to handle such preprocessing in c?

#include<stdio.h>
#include<math.h>
#define sq(x) ((x<0)?sqrt(-x):sqrt(x))

int main()
{
    int x;
    x=sq(-9);
    printf("%d\n",x);
    return 0;
}

Answer 1

because your # define "sq" checks if its a negative number and turns it into a positive number before calculating a square root

its doing sqrt(-x) which is sqrt(-(-9)) ( taking the negative of a negative is the positive)

so its doing sqrt(9)

Answer 2

You have this define:

define sq(x)  ((x<0)?sqrt(-x):sqrt(x))

Since you're passing -9, x<0 is true, so it's doing sqrt(9), which is 3.

You then print the 3.

Answer 3

The code is doing exactly what it's being told, I think.

sq(x) tests first for x < 0 - in your case, this is true, so sq(x) calls sqrt(-(-9)) - which is 3.

An attempted solution is (x < 0) ? -sqrt(-x) : sqrt(x), which will return negative roots for negative x (my best interpretation of your intent).

Answer 4

Wait, wait, wait.

Are we trying to break the basic rules of Math here?

The square root of -9 is 3 i.

That's because (-3)^2 is 9. Negative numbers have 'imaginary' square roots. sqrt(-1) is i. Not -1. (-1)^2 is 1.

Answer 5

If you expect -3, then you probably meant

#define sq(x)  ((x<0)?-sqrt(-x):sqrt(x))

but why would you expect negative square root of negative number is beyond me.