c++ - 在迭代器的计算中，如果我使用 begin() 之前或 end() 之后的迭代器作为操作数，它的行为是否定义？

Question

我做了一个实验:

#include <iostream>
#include <vector>

int main(void) {
  std::vector<int> a{1, 2, 3};
  std::vector<int>::iterator b = a.begin();
  std::vector<int>::iterator c = a.end();
  std::vector<int>::iterator d = b - 1;
  std::vector<int>::iterator e = c + 1;
  std::cout << true << std::endl;
  std::cout << (d < b) << std::endl;
  std::cout << (e > c) << std::endl;
  return 0;
}

输出:

1
1
1

但是有人告诉我deque是未定义的行为，你怎么看？谢谢!

Answer 1

没有。

d 和 e 都是“单数迭代器”，因为它们既不指向序列中的元素，也不指向“尾后一个”伪-元素。

而且你几乎不能用单数迭代器做任何事情:

[C++11: 24.2.1/5]: Just as a regular pointer to an array guarantees that there is a pointer value pointing past the last element of the array, so for any iterator type there is an iterator value that points past the last element of a corresponding sequence. These values are called past-the-end values. Values of an iterator i for which the expression *i is defined are called dereferenceable. The library never assumes that past-the-end values are dereferenceable. Iterators can also have singular values that are not associated with any sequence. [ Example: After the declaration of an uninitialized pointer x (as with int* x;), x must always be assumed to have a singular value of a pointer. —end example ] Results of most expressions are undefined for singular values; the only exceptions are destroying an iterator that holds a singular value, the assignment of a non-singular value to an iterator that holds a singular value, and, for iterators that satisfy the DefaultConstructible requirements, using a value-initialized iterator as the source of a copy or move operation.

请注意，您不能对它们进行任意比较。

我很确定即使评估 a.begin() - 1 或 a.end() + 1 也是 UB，但我找不到任何证据现在。