Operator overload question

This is a discussion on Operator overload question within the C++ Programming forums, part of the General Programming Boards category; I know that for + and - you use the following code; Code: complexClass complexClass::operator+ (complexClass a){} But can I ...

  1. #1
    843
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    Operator overload question

    I know that for + and - you use the following code;
    Code:
    complexClass complexClass::operator+ (complexClass a){}
    But can I use similar codes for *, /, and ^? Are they the same?
    Last edited by 843; 03-13-2011 at 11:52 AM.

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    Yes.
    Pass by const reference:
    Code:
    complexClass complexClass::operator+ (const complexClass& a){}
    I never put signature, but I decided to make an exception.

  3. #3
    843
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    Why do I need const and to pass by reference? Why don't the + and - operators need to?

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    Algorithm Dissector iMalc's Avatar
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    Quote Originally Posted by kmdv View Post
    Yes.
    Pass by const reference:
    Code:
    complexClass complexClass::operator+ (const complexClass& a){}
    no that's not right, if you use the one-parameter member function version then it should also be a const function.
    However, for the operators that return a result by value, e.g. +, -, *, /, |, ^, & etc, you should use the two-parameter friend, static or non-member version.
    e.g.
    Code:
    friend complexClass complexClass::operator+ (const complexClass &a, const complexClass &b)
    {
        complexClass c;
        // ... blah blah ...
        return c;
    }
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    C++まいる!Cをこわせ! Elysia's Avatar
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    Quote Originally Posted by 843 View Post
    Why do I need const and to pass by reference? Why don't the + and - operators need to?
    All operators should. And the why is because it's more efficient. You don't need to make copies of the objects all the time.
    And the const simply stops you from accidentally modifying the objects you are trying to work with.
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    You mean it's included as a crutch to help ancient programmers limp along without them having to relearn too much.

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  6. #6
    843
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    I see now. Thanks for all the help!

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