To the power of in C? [duplicate]

Solution 1:

You need pow(); function from math.h header.
syntax

#include <math.h>
double pow(double x, double y);
float powf(float x, float y);
long double powl(long double x, long double y);

Here x is base and y is exponent. result is x^y.

usage

pow(2,4);  

result is 2^4 = 16. //this is math notation only   
// In c ^ is a bitwise operator

And make sure you include math.h to avoid warning ("incompatible implicit declaration of built in function 'pow' ").

Link math library by using -lm while compiling. This is dependent on Your environment.
For example if you use Windows it's not required to do so, but it is in UNIX based systems.

Solution 2:

you can use pow(base, exponent) from #include <math.h>

or create your own:

int myPow(int x,int n)
{
    int i; /* Variable used in loop counter */
    int number = 1;

    for (i = 0; i < n; ++i)
        number *= x;

    return(number);
}

Solution 3:

#include <math.h>


printf ("%d", (int) pow (3, 4));

Solution 4:

There's no operator for such usage in C, but a family of functions:

double pow (double base , double exponent);
float powf (float base  , float exponent);
long double powl (long double base, long double exponent);

Note that the later two are only part of standard C since C99.

If you get a warning like:

"incompatible implicit declaration of built in function 'pow' "

That's because you forgot #include <math.h>.