stl question

Stl''s sort uses the operator < to sort the values. You can either change to a class and overload it as a member function or overload it like this

bool operator<(const Node & n1, const Node & n2) { return (n1.f<n2.f); }

You need to either overload comparison operators for your new types or define an appropriate binary predicate that you then pass to sort.

Like

bool 
OrderByH( const Node& lhs, const Node& rhs )
{
  return lhs.h < rhs.h;
}

...

sort( List.begin(), List.end(), OrderByH );
      

quote:

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Original post by hewhay
Stl''s sort uses the operator < to sort the values. You can either change to a class and overload it as a member function or overload it like this

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No, that''s the _default_ behaviour. It may be indesirable to define a ''global'' ordering criterion for a given type (e.g. complex numbers), while still being able to sort them in some cases (by real part or imaginary part).

Instead of the binary predicate, it may be faster to use list''s own sort() function, and pass it a function object. That''s how I''d do this, anyway.

Chris Barry (crbarry at mts.net)
My Personal Programming Depot

You''re right, you must use list''s sort() member function.
But the parameter isn''t necessarily a function object, it can be a binary predicate (i.e. a function taking 2 parameters and returning a bool).

PSA (Public Service Announcement):
std::list is the worst container you can sort. That''s why it has its own sort member function (rather than std::sort, which can be used on vectors and dequeus; maps and sets are always sorted, of course). If you''re sorting once, it''s probably OK. If you''re sorting a lot, you might want to look into a different container.