Alex Rivera | Logout

How do conversion operators work in C++?

Asked 2009-08-20T18:03:26.173
58

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?

c++ conversion-operator
Edit
Report

2 Answers

8

Implicit conversion (whether by conversion operators or non-explicit constructors) occurs when passing parameters to functions (including overloaded and default operators for classes). In addition to this, there are some implicit conversions performed on arithmetic types (so adding a char and a long results in the addition of two longs, with a long result).

Implicit conversion does not apply to the object on which a member function call is made: for the purposes of implicit conversion, "this" is not a function parameter.

answered 2009-08-20T18:19:58.137
2

The compiler will attempt one(!) user-defined cast (implicit ctor or cast operator) if you try to use an object (reference) of type T where U is required.

The . operator, however, will always try to access a member of the object (reference) on its left side. That's just the way it's defined. If you want something more fancy, that's what operator->() can be overloaded for.

answered 2009-08-20T18:19:04.473

Your Answer