Overloading by return type
I read few questions here on SO about this topic which seems yet confusing to me. I've just begun to learn C++ and I haven't studied templates yet or operator overloading and such.
Now is there a simple way to overload
class My {
public:
int get(int);
char get(int);
}
without templates or strange behavior? or should I just
class My {
public:
int get_int(int);
char get_char(int);
}
?
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.
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, you can't overload by return type; only by parameter types, and const/volatile qualifiers.
One alternative would be to "return" using a reference argument:
void get(int, int&);
void get(int, char&);
although I would probably either use a template, or differently-named functions like your second example.