C ++ STL：将空容器传递给lower_bound

Question

Is the behavior for passing an empty container to std::lower_bound defined?

I checked cppreference.com and an old version of the C++ standard that I found online, but couldn't find a definite answer.

The cppreference.com documentation for std::deque::erase has a sentence

The iterator first does not need to be dereferenceable if first==last : erasing an empty range is a no-op.

I miss something like this for std::lower_bound and other algorithms.

Answer 1

Cppreference on the return value of std::lower_bound(first, last) :

"[it returns] Iterator pointing to the first element that is not less than value, or last if no such element is found . ".

(emphasis mine)

In an empty range, there will be no elements that satisfy the criteria, so last will be returned.

Concluding from this, applying std::lower_bound (and similar) on the empty range is well-defined . It does nothing and returns last , which is equal to first .

Answer 2

The Standard says :

Returns: The furthermost iterator i in the range [first, last] such that for every iterator j in the range [first, i) the following corresponding conditions hold: *j < value or comp(*j, value) != false .

Now:

1. The range [first, last] for an empty container has a single member, namely the iterator returned by its member function end() .
2. i can therefore by only end() .
3. There is only one viable range [first, i) , which is [end, end()) .
4. This range is empty, since there is no element that is greater or equal than end() and lower then end() at the same time.

Since there is no every iterator j , I guess the quoted sentence can be rewritten into:

Returns: The furthermost iterator i in the range [first, last] .

Which implies that the only i that can be returned is end() .

Answer 3

std :: lower_bound回答了这个问题，“Where（就像在迭代器值之前）是第一个可以插入给定元素而不违反排序的地方？”如果给定的[ first，last ]范围为空，则只有在最后一个位置插入任何东西（等于第一个 ），所以这就是lower_bound返回的内容。