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c++sortingstloperator-overloadingfunctor
Andy
Andy

Reputation: 2830

Defining < for STL sort algorithm - operator overload, functor or standalone function?

I have a stl::list containing Widget class objects. They need to be sorted according to two members in the Widget class.

For the sorting to work, a less-than comparator comparing two Widget objects must be defined. There seems to be a myriad of ways to do it. From what I can gather, one can either:

a. Define a comparison operator overload in the class:

bool Widget::operator< (const Widget &rhs) const

b. Define a standalone function taking two Widgets:

bool operator<(const Widget& lhs, const Widget& rhs);

And then make the Widget class a friend of it:

class Widget {
    // Various class definitions ...
    friend bool operator<(const Widget& lhs, const Widget& rhs);
};

c. Define a functor and then include it as a parameter when calling the sort function:

class Widget_Less :
public binary_function<Widget, Widget, bool> { 
    bool operator()(const Widget &lhs, const Widget& rhs) const;
};

Does anybody know which method is better? In particular I am interested to know if I should do 1 or 2. I searched the book Effective STL by Scott Meyer but unfortunately it does not have anything to say about this.

Thank you for your reply.

Upvotes: 10

Views: 17652

Answers (4)

rlbond
rlbond

Reputation: 67789

For most purposes, a. and b. are the same. So the real question is, when to use a/b and when to use c.

The answer is: use a or b if "less than" makes sense for your object in unequivocal terms. If your class is a number, use <.

If "less than" makes no sense in the context of your class, then please don't overload "operator<" for your class. It will confuse users. Use c. instead, and either make it a nested class or typedef it inside your class so you can write it as Widget::Compare.

Upvotes: 1

bayda
bayda

Reputation: 13581

a. b. Comparison operator for two Widgets is not intuitive thing as for me. Now I can't see what it can do. Also if this function is not intuitive once you will need one new comparison operator, what you can do in this case?

I prefer functor.

Upvotes: 2

SoapBox
SoapBox

Reputation: 20609

If you are only comparing two Widgets to each other, use a member operator <. If you are comparing Widget to something else, define a global operator < (the two parameter version, optionally a friend of the Widget class but that is a separate issue.

Functor you really only want if you are doing something a little less orthodox. Choose a functor if a "less than" comparison doesn't make sense in the context of widgets. In that case, having operator < could be confusing. Of course, functors still have to provide an ordering, but just because it is an ordering doesn't really mean it is a "less than" operation. (Example, sorting states by population is probably better for a functor than an operator <.

Upvotes: 11

Peter Alexander
Peter Alexander

Reputation: 54270

They should all be the same in terms of performance, but there are other differences between them:

  • The first two save you having to explicitly specify the comparator, and can be used easily with other operations, possibly poorly defined ones that don't allow explicit specification of a comparator.

  • Only the functor allows additional data for the comparison. For example, if you were comparing ints, you could create a comparison that compares their distance from a third point, P, which would be a member of the functor instance.

  • Functors are generally less easy to read (to those not familiar with C++).

Note, you don't need to inherit binary_operator for it to work, although it does give you some nice typedefs.

Upvotes: 1

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