[boost] Re: lexicographic_less

From: jsiek_at_[hidden]
Date: 2000-06-08 13:24:17

I like using std::pair for solving this problem. Just use std::pair
like a lisp-style linked-list, and the operator< for std::pair gives
you lexicographical compare for free.

If you need members of type A, B, C, and D then create the list:

pair<A, pair<B, pair<C,D> >

or if you want to use a nil class:

pair<A, pair<B, pair<C, pair<D, nil> > >

Cheers,

Jeremy

Dave Abrahams writes:
 > --- In boost_at_[hidden], "Dave Abrahams" <abrahams_at_m...> wrote:
 > > How often have you written an aggregate class from which you needed 
 > > to use all or some data members to form a sort key for a standard 
 > > container or algorithm? Happens to me all the time. As such, I 
 > > propose the following suite of template functions as an addition to 
 > > utility.hpp (please, someone, suggest a better name!)
 > 
 > That'll teach me to post before testing; my code doesn't work. I 
 > still think it would be good to find a way to address this problem, 
 > but I am less convinced I have the right way, even after I fix the 
 > bugs. Ultimately, one doesn't want a family of template functions, 
 > but of classes (so that they can be used in the standard associative 
 > containers). Also, it would be good to be able to specify the 
 > comparison criterion for each sub-comonent.
 > 
 > -Dave
 > 
 > 
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