Question

Is there a defined behavior for container.erase(first,last) when first == last in the STL, or is it undefined?

Example:

std::vector<int> v(1,1);
v.erase(v.begin(),v.begin());
std::cout << v.size(); // 1 or 0?

If there is a Standard Library specification document that has this information I would appreciate a reference to it.

Answer 1

The behavior is well defined.

It is a No-op(No-Operation). It does not perform any erase operation on the container as end is same as begin.

The relevant Quote from the Standard are as follows:

C++03 Standard: 24.1 Iterator requirements and
C++11 Standard: 24.2.1 Iterator requirements

Para 6 & 7 for both:

An iterator j is called reachable from an iterator i if and only if there is a finite sequence of applications of the expression ++i that makes i == j. If j is reachable from i, they refer to the same container.

Most of the library’s algorithmic templates that operate on data structures have interfaces that use ranges.A range is a pair of iterators that designate the beginning and end of the computation. A range [i, i) is an empty range; in general, a range [i, j) refers to the elements in the data structure starting with the one pointed to by i and up to but not including the one pointed to by j. Range [i, j) is valid if and only if j is reachable from i. The result of the application of functions in the library to invalid ranges is undefined.