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

Alberto Ganesh Barbati <>
Thu, 3 Apr 2008 15:07:40 CST
andreas ha scritto:

I have seen a piece of code where the key of a std::map is like:

struct key
    int a;
    int b;

The less operator is defined like:

operator<(const key &lhs, const key &rhs)
    if (lhs.a == 0 || rhs.a == 0)
        return false;

    if (lhs.a != rhs.a)
        return lhs.a < rhs.a;

    if (lhs.b == 0 || rhs.b == 0)
        return false;

    if (lhs.b != rhs.b)
        return lhs.b < rhs.b;

    return false;

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.

There are only elements inserted where a and b are != 0.

It doesn't matter which elements are in the map. The behaviour is
undefined anyway.

Afterwards equal_range is called with a key where a == 0 or where (a
== 0 and b == 0). So
equal_range for the std::map returns more than one elements.

This works fine. But is this defined behaviour?

As I said it before, no.

In normal situations, that is if the comparator induces a Strict Weak
Order, std::map<>::equal_range will always return at most one element.



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