overloaded operators

This is a discussion on overloaded operators within the C++ Programming forums, part of the General Programming Boards category; what is difference in these two codes as they are giving different outputs..?? Code: circle operator=(circle c) { circle t; ...

  1. #1
    Registered User
    Join Date
    Oct 2012
    Posts
    8

    overloaded operators

    what is difference in these two codes as they are giving different outputs..??
    Code:
    circle operator=(circle c)
    {
    circle t;
    t.radius=c.radius;
    t.x=c.x;
    t.y=c.y;
    return t;
    }
    Code:
    circle operator=(circle c)
    {
    radius=c.radius;
    x=c.x;
    y=c.y;
    return circle(radius,x,y);
    }

  2. #2
    Registered User
    Join Date
    Jun 2005
    Posts
    6,628
    I am assuming both of those function definitions are within the definition for class circle. Otherwise, both samples will give a compilation error.

    Given an expression "lhs = rhs;" (where lhs and rhs are both of type circle) the first version does not change lhs. The second one does.

    Conventionally, assignment operators accept an argument that is a const reference, and return a reference. Rather than (as in both your cases) accepting an argument by value and returning a value.

    Conceptually, you are missing the fact that assignment operators are member functions, with an implied "this" pointer.
    Right 98% of the time, and don't care about the other 3%.

    If I seem grumpy or unhelpful in reply to you, or tell you you need to demonstrate more effort before you can expect help, it is likely you deserve it. Suck it up, Sunshine, and read this, this, and this before posting again.

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