如果只定义了一个小于比较器，upper_bound() 是如何工作的？

Question

I was given a class that is built on top of a map with the following K type and V type limitations:

• K : copyable, assignable, is less-than comparable ( < ). Does not implement any other operations (no equality comparison or arithmetic operators)

• V : copyable, assignable, equality-comparable ( == ). Does not implement any other operations.

Given that limitation, wouldn't this code not work?

auto it = map.upper_bound(K);

Since upper_bound() (as defined here ) returns an iterator pointing to the first element that is greater than key. Meaning the K would use a greater-than comparator?

Or does it follow that a definition of a less-than comparator would also define a greater-than comparator?

Or is my understanding of how upper_bound() works wrong?

Answer 1

As commented by @Nate Eldredge:

I believe you would get the first element y such that key < y. That's what "greater than" means in this context. Not the first y such that y > key

and from C++ documentation of upper_bound() :

template <class ForwardIterator, class T>
ForwardIterator upper_bound (ForwardIterator first, ForwardIterator last, const T& val)
{
 ForwardIterator it;
 iterator_traits<ForwardIterator>::difference_type count, step;
 count = std::distance(first,last);
 while (count>0)
{
    it = first; step=count/2; std::advance (it,step);
    if (!(val<*it))                 // or: if (!comp(val,*it)), for version (2)
      { first=++it; count-=step+1;  }
    else count=step;
  }
  return first;
}

upper_bound() uses a < operator for comparison or to be more specific this: !(val<*it)