return type of conversion opeator

Hello everyone,


In a class T, if there is a conversion opeator G, we always define it like this,

operator G()

without return type. So the implicit return type is G or G&? I can not find it in the C++ programming language book. :-)


thanks in advance,
George

How about trying?

Code:

---

#include <iostream>

struct A
{
    int n;
    operator int()
    {
        std::cout << "Conversion to int\n";
        return n;
    }
};

int main()
{
    A a;
    a.n = 100;
   
    //error: invalid initialization of reference of type 'int&' from expression of type 'A'
    //int& s = a;
   
    int t = a;
    t = 42;
    std::cout << a.n << ' ' << t << '\n';
    std::cin.get();
}

---

The error would go away if you provided an operator int& but then the initialization of t would become ambigous.

I would avoid creating custom conversion operators all together. It's too easy for them to get called when you don't intend for them to be called. Why not provide an explicit function to convert to a new type, kind of like std::string::c_str().

To answer the original question, operator G() has an implicit type of G.

To respond to cpjust, I agree the problem of unwanted implicit conversions can be a significant one, but disagree with that being a reason to avoid custom conversions altogether. With sensible design, it is possible to identify cases where implicit conversions are desirable (eg implicit conversions between very high precision floating point and very large integral types, should those be implemented class types).

I agree with grumpy. Some things are safe - mostly returning const of some type. But when you need a non-const, you should explicitly call a member function to attain it. This applies especially to classes that wrap around buffers and memory.