Overloading by return type

It's possible, but I'm not sure that it's a technique I'd recommend for beginners. As in other cases, when you want the choice of functions to depend on how the return value is used, you use a proxy; first define functions like getChar and getInt, then a generic get() which returns a Proxy like this:

class Proxy
{
    My const* myOwner;
public:
    Proxy( My const* owner ) : myOwner( owner ) {}
    operator int() const
    {
        return myOwner->getInt();
    }
    operator char() const
    {
        return myOwner->getChar();
    }
};

Extend it to as many types as you need.

No there isn't. You can't overload methods based on return type.

Overload resolution takes into account the function signature. A function signature is made up of:

• function name
• cv-qualifiers
• parameter types

And here's the quote:

1.3.11 signature

the information about a function that participates in overload resolution (13.3): its parameter-type-list (8.3.5) and, if the function is a class member, the cv-qualifiers (if any) on the function itself and the class in which the member function is declared. [...]

Options:

1) change the method name:

class My {
public:
    int getInt(int);
    char getChar(int);
};

2) out parameter:

class My {
public:
    void get(int, int&);
    void get(int, char&);
}

3) templates... overkill in this case.