How Do Conversion Operators Work in C++?

Question

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?

Accepted Answer

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.

Must Read

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!

How Do Conversion Operators Work in C++?

7 Answers

Must Read

Conversion during argument passing

James H. Sterling