Re: Is it allowed that std::map equal_range returns more than 1 elements?

Alberto Ganesh Barbati <>
Sat, 5 Apr 2008 13:26:53 CST
Bart van Ingen Schenau ha scritto:

Alberto Ganesh Barbati wrote:

This operator does not induce a Strict Weak Ordering, because the !(x
< y) && !(y < x) is not an equivalence relation. In particular the
(0,0) would be "equivalent" with every other key. As your operator<
does not satisfy this basic requirement, using it in a std::map
produces undefined behaviour.

Can you quote the standard on that for me?

I already did in another post. See also 25.3 for the precise definition
of strict weak ordering.

Because, if it is true, then the harmlessly looking definition
   std::set<double> s;
also results in UB. std::less<double> does not give an equivalence
relation if one of the two values is a NAN, or rather under the
equivalence rules of std::set<>, a NAN is equivalent with all other

I admit that I never thought about that. I'm not a floating point expert
but AFAIK the C++ standard does not specify floating point arithmetic in
presence of NaNs, it doesn't even mandate the presence of them. So we
could have a C++ implementation without NaNs or with signaling NaNs
only: in both cases std::less<> would induce a strict weak ordering.
However, I think you are correct: using a floating point type with quiet
NaNs in associative containers with std::less<> is a recipe for
undefined behaviour. Of course, no one will notice until a NaN is
actually inserted in the container... On the other hand, it's easy to
provide a different comparator that doesn't suffer this problem, for

   template <typename T>
   struct std::safe_less
     bool operator()(T x, T y) const
       if (isnan(x))
         return false;
         return isnan(y) ? true : x < y;

but I guess no one is going to use that...


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