exsnake - 1 year ago 50

C Question

`#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`

`double`

`25`

`24`

`25`

`n=4; n=6; n=3; n=2;`

Answer Source

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 to`pow`

.