Thread: operator overloading

  1. #16
    and the Hat of Guessing tabstop's Avatar
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    Quote Originally Posted by arian View Post
    I have another question too.

    I am confused:
    I got 2 example functions which returns "*this".
    one of them is like:
    Code:
    class_name &  fun_name (class_name);
    the other one is:
    Code:
    class_name fun_name (class_name);
    if it is "return(*this)" in both of them , how come the return type is different?!
    Both of these return a "class_name", so that's ok. The first returns by reference, while the second returns by "copy".

  2. #17
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    Quote Originally Posted by tabstop View Post
    Really? Do you not see the difference between "nothing" and "something"?
    No,

    I mean
    why --a calls [ loc operator--(); ]
    and
    why a-- calls [ loc operator--(int); ]

  3. #18
    Lurking whiteflags's Avatar
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    The function signatures for prefix and postfix are different, so if you can tell them apart, the compiler sure can. Is loc operator-- ( int ); completely the same as loc operator-- ( ); ? Now all you need to know is which does what operation.

  4. #19
    and the Hat of Guessing tabstop's Avatar
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    Quote Originally Posted by arian View Post
    No,

    I mean
    why --a calls [ loc operator--(); ]
    and
    why a-- calls [ loc operator--(int); ]
    That's how the operators are defined. --a is defined to be operator--() and a-- is defined to be operator--(int).

  5. #20
    The Dragon Reborn
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    Quote Originally Posted by arian View Post
    I have another question too.

    I am confused:
    I got 2 example functions which returns "*this".
    one of them is like:
    Code:
    class_name &  fun_name (class_name);
    the other one is:
    Code:
    class_name fun_name (class_name);
    if it is "return(*this)" in both of them , how come the return type is different?!
    I believe because it is a reference can be treated like the object itself, so the compiler wouldn't mind?
    Although if it was a pointer, it would very much mind
    You ended that sentence with a preposition...Bastard!

  6. #21
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    Quote Originally Posted by arian View Post
    I found this

    Code:
    loc loc::operator--(int)
    {
    	loc temp=*this;
                    lenght--;
    	width--;
    	return (temp);
    }
    will solve the problem.

    "this" points to the current obj.
    Before changing anything, copy it into temp.
    by
    lenght--;
    width--;
    the object will be changed but the previous value is returned by return(temp).
    I don't get it.
    You do not update the object however?
    so how will it get updated when you return temp?
    unless you do
    obj1=--obj1 :S
    You ended that sentence with a preposition...Bastard!

  7. #22
    and the Hat of Guessing tabstop's Avatar
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    Quote Originally Posted by Eman View Post
    I don't get it.
    You do not update the object however?
    That's what length-- and width-- do: they modify the object.

  8. #23
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    Quote Originally Posted by tabstop View Post
    That's what length-- and width-- do: they modify the object.
    oh right
    Last edited by Eman; 12-22-2010 at 03:17 PM.
    You ended that sentence with a preposition...Bastard!

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