functional derivation better?

This is a discussion on functional derivation better? within the C++ Programming forums, part of the General Programming Boards category; I'm told that if I derive my comparison function objects from std::binary_function, I'm "helping the STL." How? What does, std::sort(), ...

  1. #1
    Kiss the monkey. CodeMonkey's Avatar
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    functional derivation better?

    I'm told that if I derive my comparison function objects from std::binary_function, I'm "helping the STL." How? What does, std::sort(), for example, deduce from that template? Why is
    Code:
    class comp_square : public std::binary_function<double,double,bool>
    {
         bool operator() (const double & left, const double & right)
         {
                return std::pow(left,2) < std::pow(right,2);
         }
    
         // . . . . . . 
    };
    better than

    Code:
    class comp_square
    {
         bool operator() (const double & left, const double & right)
         {
                return std::pow(left,2) < std::pow(right,2);
         }
    
         // . . . . . . 
    };
    "If you tell the truth, you don't have to remember anything"
    -Mark Twain

  2. #2
    Registered User MacNilly's Avatar
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    In that example, there is no difference at all, reason is... std::binary_function is never used...
    GCC on Fedora Core 10

  3. #3
    Captain Crash brewbuck's Avatar
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    Quote Originally Posted by CodeMonkey View Post
    I'm told that if I derive my comparison function objects from std::binary_function, I'm "helping the STL." How? What does, std::sort(), for example, deduce from that template?
    It gives the typedefs: first_argument_type, second_argument_type, result_type.

    In theory, the STL (or your own algorithms even) might have some partial or complete specialization of an algorithm based on the specific types being passed to a functor. This information COULD be carried around as template type parameters, but this is unwieldy. It would force the user programmer to explicitly indicate the types for every instantiation of an object which acts like a binary function, even though the vast majority of algorithms don't make use of this information.


    You also need the result_type if you want boost::bind to work correctly with your function object. You could define this yourself (and I have), but you could just as well derive from binary_function.

  4. #4
    Kiss the monkey. CodeMonkey's Avatar
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    I see. Thank you.
    "If you tell the truth, you don't have to remember anything"
    -Mark Twain

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