Why is my power operator (^) not working?
C++CC++ Problem Overview
#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?
C++ Solutions
Solution 1 - C++
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.
Solution 2 - C++
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 );
Solution 3 - C++
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 int
s:
int power(int base, int exp)
{
int result = 1;
while(exp) { result *= base; exp--; }
return result;
}
Solution 4 - C++
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);
Solution 5 - C++
include math.h and compile with gcc test.c -lm
Solution 6 - C++
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;
}`
Solution 7 - C++
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.
Solution 8 - C++
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;
}
Solution 9 - C++
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