Why does pow(n,2) return 24 when n=5, with my compiler and OS?
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
int main()
{
int n,i,ele;
n=5;
ele=pow(n,2);
printf("%d",ele);
return 0;
}
The output is 24
.
I'm using GNU/GCC in Code::Blocks.
What is happening?
I know the pow
function returns a double
, but 25
fits an int type so why does this code print a 24
instead of a 25
? If n=4; n=6; n=3; n=2;
the code works, but with the five it doesn't.
Solution 1:
Here is what may be happening here. You should be able to confirm this by looking at your compiler's implementation of the pow
function:
Assuming you have the correct #include's, (all the previous answers and comments about this are correct -- don't take the #include
files for granted), the prototype for the standard pow
function is this:
double pow(double, double);
and you're calling pow
like this:
pow(5,2);
The pow
function goes through an algorithm (probably using logarithms), thus uses floating point functions and values to compute the power value.
The pow
function does not go through a naive "multiply the value of x a total of n times", since it has to also compute pow
using fractional exponents, and you can't compute fractional powers that way.
So more than likely, the computation of pow
using the parameters 5 and 2 resulted in a slight rounding error. When you assigned to an int
, you truncated the fractional value, thus yielding 24.
If you are using integers, you might as well write your own "intpow" or similar function that simply multiplies the value the requisite number of times. The benefits of this are:
You won't get into the situation where you may get subtle rounding errors using
pow
.Your
intpow
function will more than likely run faster than an equivalent call topow
.
Solution 2:
You want int result from a function meant for doubles.
You should perhaps use
ele=(int)(0.5 + pow(n,2));
/* ^ ^ */
/* casting and rounding */