Question

I want to consider to code. First it is:

#include <iostream>
#include <cmath>
#include <math.h>
using namespace std;
int main() {
    int s = 25;
    cout << sqrt(s) << endl;
    return 0;
}

It gave me this mistake:

>c:\users\datuashvili\documents\visual studio 2010\projects\training\training\training.cpp(9): error C2668: 'sqrt' : ambiguous call to overloaded function
1>          c:\program files\microsoft visual studio 10.0\vc\include\math.h(589): could be 'long double sqrt(long double)'
1>          c:\program files\microsoft visual studio 10.0\vc\include\math.h(541): or       'float sqrt(float)'
1>          c:\program files\microsoft visual studio 10.0\vc\include\math.h(127): or       'double sqrt(double)'
1>          while trying to match the argument list '(int)'
========== Build: 0 succeeded, 1 failed, 0 up-to-date, 0 skipped ==========

If I add type float in brackets in front of s, like this:

#include <iostream>
#include <cmath>
#include <math.h>
using namespace std;

int main() {
    int s = 25;
    cout << sqrt((float)s) << endl;
    return 0;
}

I got as I would guess, 5. And another variant is that instead of sqrt, if I write sqrtf:

#include <iostream>
#include <cmath>
#include <math.h>
using namespace std;

int main(){
    int s=25;
    cout << sqrtf((float)s) << endl;
    return 0;
}

I also got 5.

What is the difference between them? Does it means that sqrtf is same as sqrt for the float type?

Answer 1

Here’s a page on MSDN documentation for sqrt() and sqrtf(), that explains the difference:

sqrt, sqrtf

Calculates the square root.

    double sqrt(
       double x 
    );
    float sqrt(
       float x 
    );  // C++ only
    long double sqrt(
       long double x
    );  // C++ only
    float sqrtf(
       float x 
    );

Parameters

x: Nonnegative floating-point value

Remarks

C++ allows overloading, so users can call overloads of
sqrt that take float or long double types. In a C program, sqrt always
takes and returns double.

Return Value

The sqrt function returns the square-root of x. If x
is negative, sqrt returns an indefinite, by default.

So the difference in C++ is that sqrt() accepts either a double, a float or a long double while sqrtf() accepts only a float.

As the documentation says, the only reason why there’s two different versions is because C did not support overloading, so there had to be two functions. C++ allows overloading, so there are actually three different versions of sqrt() taking floating point arguments of various sizes.

So in C++, both your code snippets do essentially the same thing. On C, however, there would’ve been a conversion from float to double in the sqrt() call.