Stl Set and struct comparison

This is a discussion on Stl Set and struct comparison within the C++ Programming forums, part of the General Programming Boards category; I have a struct: Code: struct aCoord { int a,b; }; and a set of them: Code: std::set<aCoord> myCoords; In ...

  1. #1
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    Stl Set and struct comparison

    I have a struct:

    Code:
    struct aCoord {
    	int a,b;
    };
    and a set of them:

    Code:
    std::set<aCoord> myCoords;
    In one of my functions, I would like to check whether a newly generated aCoord is equivalent to one of the elements already in myCoords (meaning check whether firstCoord.a==secondCoord.a && firstCoord.b == secondCoord.b).

    I have an iterator and am doing the following:

    Code:
    iterator = myCoords.find(newCoord);
    if(iterator == myCoords.end()) {
        myCoords.insert(newCoord);
    }
    This doesn't work because I need to overload the comparison function used by find, but despite the research I've done so far, I can't figure out the proper form of the operator override. Can anyone help me out? Thanks.

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    I may be wrong, but I think that find() simply uses operator< to determine equivalence (if a is not < b and b is not < a, a == b). Try defining operator< for aCoord.
    If I did your homework for you, then you might pass your class without learning how to write a program like this. Then you might graduate and get your degree without learning how to write a program like this. You might become a professional programmer without knowing how to write a program like this. Someday you might work on a project with me without knowing how to write a program like this. Then I would have to do you serious bodily harm. - Jack Klein

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    That's the part I can't figure out how to write. What is the form for that? Thanks

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    It depends on whether you write it as a member function or not. As a member function:
    Code:
    bool operator<(const aCoord&) const;
    As a regular function:
    Code:
    bool operator<(const aCoord&, const aCoord&);
    If I did your homework for you, then you might pass your class without learning how to write a program like this. Then you might graduate and get your degree without learning how to write a program like this. You might become a professional programmer without knowing how to write a program like this. Someday you might work on a project with me without knowing how to write a program like this. Then I would have to do you serious bodily harm. - Jack Klein

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    Quote Originally Posted by pollypocket4eva View Post
    That's the part I can't figure out how to write. What is the form for that? Thanks
    Note that a set already checks for uniqueness for you - you don't have to do this manually.

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    Quote Originally Posted by EVOEx View Post
    Note that a set already checks for uniqueness for you - you don't have to do this manually.
    Good catch. I keep forgetting that insert() will gracefully handle duplicates. You can check the return value of insert() to find out if a new element was actually added to the set.
    If I did your homework for you, then you might pass your class without learning how to write a program like this. Then you might graduate and get your degree without learning how to write a program like this. You might become a professional programmer without knowing how to write a program like this. Someday you might work on a project with me without knowing how to write a program like this. Then I would have to do you serious bodily harm. - Jack Klein

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    I'm not sure I understand what you're doing with the member and regular functions.
    Where would I declare those? And how do I check the components (firstCoord.a==secondCoord.a && firstCoord.b == secondCoord.b) or do I not have to do that explicitly?

    Also, EVOEx, that's very helpful to know, but in this case I want to explicitly know whether the set already contains a particular item.

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    Oh, so will insert() return false if the element was not added?

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    The insert() approach is not working for me, I get the error: No match for 'operator<'

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    Yes, that's because you must define some sort of comparison operation for the elements of a set. You may want to read the operator overloading tutorial.
    If I did your homework for you, then you might pass your class without learning how to write a program like this. Then you might graduate and get your degree without learning how to write a program like this. You might become a professional programmer without knowing how to write a program like this. Someday you might work on a project with me without knowing how to write a program like this. Then I would have to do you serious bodily harm. - Jack Klein

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    I've read that, but can't figure out how to apply it here.
    Would it be something like this?
    Code:
    struct aCoord {
    	int a,b;
    	bool operator==(const aCoord &other) const
    		{ return (a == other.a)&&(b == other.b); }
    		
    };

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    Anyone?

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    Something like that, but not that. You need operator<, not operator==. sets determine equivalence (operator==) from comparison.
    If I did your homework for you, then you might pass your class without learning how to write a program like this. Then you might graduate and get your degree without learning how to write a program like this. You might become a professional programmer without knowing how to write a program like this. Someday you might work on a project with me without knowing how to write a program like this. Then I would have to do you serious bodily harm. - Jack Klein

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    Ok, but how can the < operator address my struct? I need to test if first.a==second.a && first.b==second.b
    The following does not do that:

    Code:
    bool operator<(const aCoord &other) const
    		{ return (a < other.a)&&(b < other.b); }
    I can't think of how to structure it.
    Thanks very much for your help.

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    Quote Originally Posted by pianorain View Post
    I think that find() simply uses operator< to determine equivalence (if a is not < b and b is not < a, a == b).
    Quote Originally Posted by pianorain View Post
    sets determine equivalence (operator==) from comparison.
    I'll put it another way. Take any two numbers (A and B). They can be the same number, or they can be different. Now, either:
    1. A < B
    2. B < A
    3. !(A < B) && !(B < A)
    Note that case 3 is the same as A == B. Do you see how every case, including operator==, uses operator<? You just have to make sure that operator< orders elements correctly.

    In cases like this, you usually want to use a lexicographical order. In other words, first sort by A, and then sort by B.
    If I did your homework for you, then you might pass your class without learning how to write a program like this. Then you might graduate and get your degree without learning how to write a program like this. You might become a professional programmer without knowing how to write a program like this. Someday you might work on a project with me without knowing how to write a program like this. Then I would have to do you serious bodily harm. - Jack Klein

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