How pow function works in this case?

Question

#include <stdio.h>
#include <math.h>

int main()
{
int i = 11;
printf("%d ^ 2 = %d\n",i,(int)pow(i,2));
getchar();
return 0;
}

In this case instead of getting 121,i am getting 120.What is the mistake i am making? (I really need to print pow(i,2) as an int.)

Why use floating point `math.h` if you need `int` arithmetics? — mouviciel, Apr 11 '14 at 12:42
Although there are, _in general_, potential rounding issues, a decent `pow()` would return the expected 121. OP has a [Schlocky](http://www.merriam-webster.com/dictionary/schlock) math library. — chux - Reinstate Monica, Apr 11 '14 at 14:44

score 5 · Answer 1 · answered Apr 11 '14 at 12:42

5

Casting to integer truncates the fraction, possibly pow returned something like 120.99999998 or so...

Don't cast to (int) and use %g format instead of %d to print double result of pow().

answered Apr 11 '14 at 12:42

CiaPan

On gcc 4.1.2, I still get 121 with those changes. Not sure what compiler OP is using – Engineer2021 Apr 11 '14 at 12:47

score 1 · Answer 2 · answered Apr 11 '14 at 12:44

1

It is to do with rounding from double to int.

As the power is constant why have the overhead.

Stick to

printf("%d ^ 2 = %d\n",i, i*i);

answered Apr 11 '14 at 12:44

Ed Heal

2 Answers2