Why does sqrt() work fine on an int variable if it is not defined for an int?

Question

In chapter 3 of Programming: Principles and Practice using C++ (sixth printing), Stroustrup states (p.68): "Note that sqrt() is not defined for an int".

Here is a simple C++ program based on that chapter:

#include "std_lib_facilities.h"

int main()
{
    int n = 3;
    cout << "Square root of n == " << sqrt(n) << "\n";
}

Given the quote above, I would expect the process of compiling or running this program to fail in some way.

To my surprise, compiling it (with g++ (GCC) 4.2.1) and running it succeeded without errors or warnings, and produced the following perfectly decent output:

Square root of n == 1.73205

My question therefore is: if sqrt() really is not defined for an int, then why doesn't the program above fail somehow?

Answer 1

The 10 is being implicitly converted to a double. This will happen automatically as long as you have the correct function prototype for sqrt.

Edit: beaten by comments