calling std::minus can't work after overloading - operator - c++

Question

Am getting compilation errors for the following snippet of my C++ code.
struct Power{
int power;
int age;
int operator-(const Power& p1)
{
return this->power - p1.power;
}
};
int main() {
Power p1;
p1.power = 1;
p1.age = 25;
Power p2;
p2.power = 2;
p2.age = 26;
std::cout<<std::minus<Power>()(p1, p2)<<std::endl;
}
build with c++11. cannot be built.
error messages are:
error: no match for ‘operator<<’ (operand types are ‘std::ostream {aka std::basic_ostream<char>}’ and ‘Power’)
std::cout<<std::minus<Power>()(p1, p2)<<std::endl;
^
In file included from /usr/include/c++/5/iostream:39:0,
from rvaluereference.cpp:1:
/usr/include/c++/5/ostream:628:5: note: candidate: std::basic_ostream<_CharT, _Traits>& std::operator<<(std::basic_ostream<_CharT, _Traits>&&, const _Tp&) [with _CharT = char; _Traits = std::char_traits<char>; _Tp = Power] <near match>
operator<<(basic_ostream<_CharT, _Traits>&& __os, const _Tp& __x)
^
/usr/include/c++/5/ostream:628:5: note: conversion of argument 1 would be ill-formed:
rvaluereference.cpp:60:39: error: cannot bind ‘std::ostream {aka std::basic_ostream<char>}’ lvalue to ‘std::basic_ostream<char>&&’
.....

Answer 1

std::minus has a single template parameter that defines the type of both inputs and the output; it doesn't handle type switching AFAICT. Your operator- takes Power, and returns int, but std::minus<Power> must take and return Power. The complaint is because std::minus<Power> returns Power, and ostream doesn't have an overload for operator<< that accepts Power.
As noted in the comments, if you can use C++14, std::minus<void> accepts mismatched arguments and deduces the return type, so if you can use C++14, that's another option.

Answer 2

While std::minus only can do T operator-(const T &lhs, const T &rhs), your operators do not have the same limitations. Declare the operator outside of the class:
int operator-(const Power &lhs, const Power &rhs) {
return lhs.power - rhs.power;
}
and now you can do std::cout << p1 - p2 << std::endl;