This question already has an answer here:
-
Strange behaviour of the pow function
5 answers
I was simply writing a program to calculate the power of an integer. But the output was not as expected. It worked for all the integer numbers except for the power of 5.
My code is:
#include <stdio.h>
#include <math.h>
int main(void)
{
int a,b;
printf("Enter the number.");
scanf("\n%d",&a);
b=pow(a,2);
printf("\n%d",b);
}
The output is something like this:
"Enter the number. 2
4
"Enter the number. 5
24
"Enter the number. 4
16
"Enter the number. 10
99
Can't we use pow()
function for int data type??
Floating point precision is doing its job here. The actual working of pow
is using log
pow(a, 2) ==> exp(log(a) * 2)
Look at math.h
library which says:
<math.h>
/* Excess precision when using a 64-bit mantissa for FPU math ops can
cause unexpected results with some of the MSVCRT math functions. For
example, unless the function return value is stored (truncating to
53-bit mantissa), calls to pow with both x and y as integral values
sometimes produce a non-integral result. ... */
Just add 0.5
to the return value of pow
and then convert it to int
.
b = (int)(pow(a,2) + 0.5);
So, the answer to your question
Does pow() work for int
data type in C?
Not always. For integer exponentiation you could implement your own function (this will work for +ve integers only):
int int_pow(int base, int exp)
{
int result = 1;
while (exp)
{
if (exp & 1)
result *= base;
exp /= 2;
base *= base;
}
return result;
}
there is no int based pow. What you are suffering from is floating point truncation.
an int based pow is too constrained (the range of inputs would quickly overflow an int). In many cases int based pow, like in your case where its powers of 2 can be done efficiently other ways.
The C library function double pow(double x, double y)
It takes double type
printf("%a", pow(10, 2))
and see what you get; I expect you'll see you don't quite get 100. Call lround
if you want to round instead of truncating.