How does the STL map::find function work without the equality operator?

问题

Under the hood, an STL map is a red-black tree, and it uses the < operator of its keys or a user-provided comparison to figure out the location for element insertion.

map::find() returns the element that matches the supplied key (if any matches are present)

How can it do this without using an equality operator? Let's say my map has the keys 1, 2, 3, and 4 in it. Using only <, I could see that the key 2 should go after 1, after 2, and before 3. But I can't tell whether or not 2 is the same as 2.

I can even see in /usr/include/c++/4.4.3/bits/stl_tree.h that find() uses nothing but the user-supplied comparison function:

template<typename _Key, typename _Val, typename _KeyOfValue,
       typename _Compare, typename _Alloc>
typename _Rb_tree<_Key, _Val, _KeyOfValue,
          _Compare, _Alloc>::iterator
_Rb_tree<_Key, _Val, _KeyOfValue, _Compare, _Alloc>::
find(const _Key& __k)
{
  iterator __j = _M_lower_bound(_M_begin(), _M_end(), __k);
  return (__j == end()
      || _M_impl._M_key_compare(__k,
                _S_key(__j._M_node))) ? end() : __j;
}

Cryptic. Bonus points if you can tell me how my comparison function ends up being used in _M_impl._M_key_compare without an obvious loop.

回答1:

If (a < b) is false and (b < a) is false, then (a == b). This is how STL's find() works.

回答2:

It uses !(a<b) && !(b<a)

来源：https://stackoverflow.com/questions/3240718/how-does-the-stl-mapfind-function-work-without-the-equality-operator