problem with CONST and function member overloading operator

[Vaaal]

Hi guys, I have a problem about const and member function overloading operator.
Well, if I do something like

int c; /c is global
const int &try (int a, int b)
{
    c=a*b;
    return c;
}

i won't be able to do
try(A,B)=5
becuase, obviously, funciont try returns a CONST int.

BUT
when I overload operator, like =, returning an object reference and declaring the member function as const, I CAN make an assignment. For instance, the member function is

const Array &Array::operator=(const Array& Ar)
{
     if (this==&Ar) cout <<"You cant' initialize an object with itself!" <<endl;
    else
    {
        cout <<"Initialization of object with an object" <<endl;
        delete [] this->ptr;
        ptr= new int[Ar.size];
        size=Ar.size;
        for (int i=0; i<size; i++)
        {
            ptr[i]=Ar.ptr[i];
        }
    }
 return *this;   
}

and in the main I do

object1=object2=object3

Why the compiler allow me to do this? I cant' figure it out. Can you help me please?

Really thank you, and sorry for my poor english.

[kbw]

object1=object2=object3;
means:

object1.operator=(
    object2.operator=(
        object3
    )
);

BTW, the compiler generates object& operator=(const object &) for copy, not with const as you have used.

[Vaaal]

BTW, the compiler generates object& operator=(const object &) for copy, not with const as you have used.

why??
So is equal to write const object & or just object &? And this is just for the operator =, or for all operator?

The deitel manual put const on an operator= function, and i'm pretty confused...

[kbw]

This applies to the assignment operator. You do realise the an assignment operator is generated for you right?

That's the way the assignment operator works. Consider this C example:

#include <stdio.h>

int main()
{
        int a = 3;
        int b, c;
        ++(c = b = a);
        printf("a=%d b=%d c=%d\n", a, b, c);

        return 0;
}

[Vaaal]

ook, so i'm wonder why the deitel uses to write const object & operator= :-\ Do you have a clue?

[hanst99]

It's a const reference so you can do this:
a=b=c;
but NOT this:
(a=b)=c;

[1yR4jE8b]

The const keyword in the context of the OP is that the reference that is returned is const. Meaning you can assign to a non-const object, but the returned reference is a const one. This is why the post above me is correct.

I think you are confusing const return values with const functions. Functions with const return values use const at the beginning in the function (as part of the return type), but const functions (functions that can be called using const instances of an object) need const at the end of the function signature.

struct Foo
{
  void foo() const; //can be called with a const Foo as well as a non-const
  void bar(); //can only be called with a non-const Foo
};

[Vaaal]

ok, now it's all clear. Thank you very much :-)