Consider this simple example:
template <class Type>
class smartref {
public:
smartref() : data(new Type) { }
operator Type&(){ return *data; }
private:
Type* data;
};
class person {
public:
void think() { std::cout << "I am thinking"; }
};
int main() {
smartref<person> p;
p.think(); // why does not the compiler try substituting Type&?
}
How do conversion operators work in C++? (i.e) when does the compiler try substituting the type defined after the conversion operator?
7 Answers
Some random situations where conversion functions are used and not used follow.
First, note that conversion functions are never used to convert to the same class type or to a base class type.
Conversion during argument passing
Conversion during argument passing will use the rules for copy initialization. These rules just consider any conversion function, disregarding of whether converting to a reference or not.
struct B { };
struct A {
operator B() { return B(); }
};
void f(B);
int main() { f(A()); } // called!
Argument passing is just one context of copy initialization. Another is the "pure" form using the copy initialization syntax
B b = A(); // called!
