The type of *this

[George2]

Hello everyone,


I am confused what is the type of *this? Type of reference to instance or type of instance?

Looks like both code can compile, no warning messages. Any ideas?

class A{
 
    A& foo1()
    {
        return *this;
    }
 
    A foo2()
    {
        return *this;
    }
};
 

thanks in advance,
George

[sagnik`]

Hello everyone,


I am confused what is the type of *this? Type of reference to instance or type of instance?

Looks like both code can compile, no warning messages. Any ideas?

class A{
 
    A& foo1()
    {
        return *this;
    }
 
    A foo2()
    {
        return *this;
    }
};
 

thanks in advance,
George

As far as I think, when you are thrying the first code, i.e.,

A& foo1()
    {
        return *this;
    }
 

the original object is returned (with updatations if assigned), but for the second case, i.e.,

A foo1()
    {
        return *this;
    }
 

a copy of the original object is returned. The compiler will not complain because return by value is acceptable, but it will undoubtedly waste some memory space.
I haven't tried this at home, but it is true that is any other case if you try to return by address in the function body and do otherwise in function declaration, the compiler gives you an error!

Let's see what other people have to say on this....

Sagnik

[George2]

Thanks Sagnik,


What does your below statement mean? Return address and do otherwise in declaration? Could you show me some pseudo code please?

but it is true that is any other case if you try to return by address in the function body and do otherwise in function declaration, the compiler gives you an error!

Sagnik

regards,
George

[mac11]

I am confused what is the type of *this? Type of reference to instance or type of instance?

this is a pointer - the return type from your function prototype tells the compiler what you want to return

When you say you're going to return a reference you return a reference, when you say you're going to return a copy, you return a copy. The compiler decides based on what your code says its going to do.

[George2]

Thanks mac11,


My question is answered.

this is a pointer - the return type from your function prototype tells the compiler what you want to return

When you say you're going to return a reference you return a reference, when you say you're going to return a copy, you return a copy. The compiler decides based on what your code says its going to do.

regards,
George