#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?
9 Answers
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.
In C ^ is the bitwise XOR:
0101 ^ 1100 = 1001 // in binaryThere's no operator for power, you'll need to use pow function from math.h (or some other similar function):
result = pow( a, i );pow() doesn't work with int, hence the error "error C2668:'pow': ambiguous call to overloaded function"
Write your own power function for ints:
int power(int base, int exp)
{ int result = 1; while(exp) { result *= base; exp--; } return result;
}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. xy
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);include math.h and compile with gcc test.c -lm
5It'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; }`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.
2There 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;
} 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.
2 << 7
