Problem with overloading an operator.

This is a discussion on Problem with overloading an operator. within the C++ Programming forums, part of the General Programming Boards category; Hello, I have to overload "<" and ">" operators. So I do: Code: bool bignum::operator<(const bignum& a) { // here ...

  1. #1
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    Problem with overloading an operator.

    Hello,
    I have to overload "<" and ">" operators. So I do:
    Code:
    bool bignum::operator<(const bignum& a)
    {
            // here is some code comparing *this and a.
    }
    now I want to use operator "<" to overload ">":
    Code:
    bool bignum::operator>(const bignum& a)
    {
            return  a < *this;
    }
    and I get a compile error:
    main.cpp: In member function `bool bignum::operator>(const bignum&)':
    main.cpp:157: error: passing `const bignum' as `this' argument of `bool bignum::operator<(const bignum&)' discards qualifiers
    Can anyone help me ?
    --
    Regards,
    apacz

  2. #2
    C++ Witch laserlight's Avatar
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    I suggest making them friend functions, and then writing:

    Code:
    bool operator<(const bignum& lhs, const bignum& rhs)
    {
    	// here is some code comparing lhs and rhs.
    }
    
    bool operator>(const bignum& lhs, const bignum& rhs)
    {
    	return rhs < lhs;
    Making them friend functions means that expressions like (1 < x) will be allowed, assuming that int is convertible to bignum (e.g. via a non-explicit constructor that accepts an int).
    Quote Originally Posted by Bjarne Stroustrup (2000-10-14)
    I get maybe two dozen requests for help with some sort of programming or design problem every day. Most have more sense than to send me hundreds of lines of code. If they do, I ask them to find the smallest example that exhibits the problem and send me that. Mostly, they then find the error themselves. "Finding the smallest program that demonstrates the error" is a powerful debugging tool.
    Look up a C++ Reference and learn How To Ask Questions The Smart Way

  3. #3
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    Thanks. But there is no way to make this overload look like: bool operator<(const bignum& rhs) ? Because your overload takes two arguments - what's the real difference between those two versions ?
    --
    Regards,
    apacz

  4. #4
    C++ Witch laserlight's Avatar
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    But there is no way to make this overload look like: bool operator<(const bignum& rhs) ?
    Yes, there is. Make your member function versions const.

    Because your overload takes two arguments - what's the real difference between those two versions ?
    The useful difference is what I mentioned: the friend function version is more flexible than the member function version. Actually, since operator> is implemented in terms of operator< you do not even need to make it a friend function.
    Quote Originally Posted by Bjarne Stroustrup (2000-10-14)
    I get maybe two dozen requests for help with some sort of programming or design problem every day. Most have more sense than to send me hundreds of lines of code. If they do, I ask them to find the smallest example that exhibits the problem and send me that. Mostly, they then find the error themselves. "Finding the smallest program that demonstrates the error" is a powerful debugging tool.
    Look up a C++ Reference and learn How To Ask Questions The Smart Way

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