C++ adding friend to a template class in order to typecast
- by user1835359
I'm currently reading "Effective C++" and there is a chapter that contains code similiar to this:
template <typename T>
class Num {
public:
Num(int n) { ... }
};
template <typename T>
Num<T> operator*(const Num<T>& lhs, const Num<T>& rhs) { ... }
Num<int> n = 5 * Num<int>(10);
The book says that this won't work (and indeed it doesn't) because you can't expect the compiler to use implicit typecasting to specialize a template.
As a soluting it is suggested to use the "friend" syntax to define the function inside the class.
//It works
template <typename T>
class Num {
public:
Num(int n) { ... }
friend
Num operator*(const Num& lhs, const Num& rhs) { ... }
};
Num<int> n = 5 * Num<int>(10);
And the book suggests to use this friend-declaration thing whenever I need implicit conversion to a template class type. And it all seems to make sense.
But why can't I get the same example working with a common function, not an operator?
template <typename T>
class Num {
public:
Num(int n) { ... }
friend
void doFoo(const Num& lhs) { ... }
};
doFoo(5);
This time the compiler complaints that he can't find any 'doFoo' at all.
And if i declare the doFoo outside the class, i get the reasonable mismatched types error. Seems like the "friend ..." part is just being ignored.
So is there a problem with my understanding? What is the difference between a function and an operator in this case?
