Why is my power operator (^) not working?

Question

#include <stdio.h>

void main(void)
{
    int a;
    int result;
    int sum = 0;
    printf("Enter a number: ");
    scanf("%d", &a);
    for( int i = 1; i <= 4; i++ )
    {
        result = a ^ i;

        sum += result;
    }
    printf("%d\n", sum);
}

Why is ^ not working as the power operator?

Answer 1

Well, first off, the ^ operator in C/C++ is the bit-wise XOR. It has nothing to do with powers.

Now, regarding your problem with using the pow() function, some googling shows that casting one of the arguments to double helps:

result = (int) pow((double) a,i);

Note that I also cast the result to int as all pow() overloads return double, not int. I don't have a MS compiler available so I couldn't check the code above, though.

Since C99, there are also float and long double functions called powf and powl respectively, if that is of any help.

Answer 2

In C ^ is the bitwise XOR:

0101 ^ 1100 = 1001 // in binary

There's no operator for power, you'll need to use pow function from math.h (or some other similar function):

result = pow( a, i );

Answer 3

pow() doesn't work with int, hence the error "error C2668:'pow': ambiguous call to overloaded function"

http://www.cplusplus.com/reference/clibrary/cmath/pow/

Write your own power function for ints:

int power(int base, int exp)
{
    int result = 1;
    while(exp) { result *= base; exp--; }
    return result;
}

Answer 4

First of all ^ is a Bitwise XOR operator not power operator.

You can use other things to find power of any number. You can use for loop to find power of any number

Here is a program to find x^y i.e. x^y

double i, x, y, pow;

x = 2;
y = 5; 
pow = 1;
for(i=1; i<=y; i++)
{
    pow = pow * x;
}

printf("2^5 = %lf", pow);

You can also simply use pow() function to find power of any number

double power, x, y;
x = 2;
y = 5;
power = pow(x, y); /* include math.h header file */

printf("2^5 = %lf", power);

Answer 5

include math.h and compile with gcc test.c -lm

Answer 6

It's not working because c as well as c++ do not have any operators to perform power operations.

What you can do is, you can use math.h library and use pow function. There is a Function for this instead of the operator.

`   #include<stdio.h>
    #include<math.h>
    int main(){
        int base = 3;
        int power = 5;
        pow(double(base), double(power));
        return 0;
     }`

Answer 7

You actually have to use pow(number, power);. Unfortunately, carats don't work as a power sign in C. Many times, if you find yourself not being able to do something from another language, its because there is a diffetent function that does it for you.

Answer 8

There is no way to use the ^ (Bitwise XOR) operator to calculate the power of a number. Therefore, in order to calculate the power of a number we have two options, either we use a while loop or the pow() function.

1. Using a while loop.

#include <stdio.h>

int main() {

    int base, expo;
    long long result = 1;

    printf("Enter a base no.: ");
    scanf("%d", &base);

    printf("Enter an exponent: ");
    scanf("%d", &expo);

    while (expo != 0) {
        result *= base;
        --expo;
    }

    printf("Answer = %lld", result);
    return 0;
}    
             

2. Using the pow() function

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

int main() {

    double base, exp, result;

    printf("Enter a base number: ");
    scanf("%lf", &base);

    printf("Enter an exponent: ");
    scanf("%lf", &exp);

    // calculate the power of our input numbers
    result = pow(base, exp);

    printf("%.1lf^%.1lf = %.2lf", base, exp, result);

    return 0;
}
     

Answer 9

If you are trying to calculate the power of base 2, you can use the bitwise shift operator to calculate the power. For example, say you wanted to calculate 2 to the power of 8.

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