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Hi guys, I know that abs() takes in an int and returns an int, while fabs() takes in a float/double and returns a float/double.
However, if I pass in an int to fabs, it works as per usual, but not when I pass a double into abs. I would expect some syntax error to appear or smth, or does type promotion occur here, where the int gets promoted to a double and so fabs() work as per usual?
In C++, std::abs is overloaded for both signed integer and floating point types. std::fabs only deals with floating point types (pre C++11). Note that the std:: is important; the C function ::abs that is commonly available for legacy reasons will only handle int!
The problem with
Copyfloat f2= fabs(-9);
is not that there is no conversion from int (the type of -9) to double, but that the compiler does not know which conversion to pick (int -> float, double, long double) since there is a std::fabs for each of those three. Your workaround explicitly tells the compiler to use the int -> double conversion, so the ambiguity goes away.
C++11 solves this by adding double fabs( Integral arg ); which will return the abs of any integer type converted to double. Apparently, this overload is also available in C++98 mode with libstdc++ and libc++.
In general, just use std::abs, it will do the right thing. (Interesting pitfall pointed out by @Shafik Yaghmour. Unsigned integer types do funny things in C++.)
With C++ 11, using abs() alone is very dangerous:
Copy#include <iostream>
#include <cmath>
int main() {
std::cout << abs(-2.5) << std::endl;
return 0;
}
This program outputs 2 as a result. (See it live)
Always use std::abs():
Copy#include <iostream>
#include <cmath>
int main() {
std::cout << std::abs(-2.5) << std::endl;
return 0;
}
This program outputs 2.5.
You can avoid the unexpected result with using namespace std; but I would adwise against it, because it is considered bad practice in general, and because you have to search for the using directive to know if abs() means the int overload or the double overload.