Why doesn't my templated function promote 'int' to 'T', where 'T' = 'double'?

Question

I have a class templated with typename T. It contains a function,

template 
myClass operator+(myClass

        

Answer 1

The compiler on overload resolution fails to find a right candidate for operator+ because T is already being deducted to double and literal 5 is an integer. Solution:

template 
myClass operator+(myClass lhs, const T2& rhs) {
    return lhs += T1(rhs);
}