C ++标准对std :: vector的评价是什么 <int> V1，V2; 性病::距离（v1.begin（），v2.begin（））？

Question

I have this code

#include <vector>
#include <iostream>

int main(int argc, char* argv[])
{
   std::vector<int> v1,v2;
   std::cout << std::distance(v1.begin(),v2.begin());
   return 0;
}

and it has a bug because it is not meaningful to compare the iterators of two different vectors.

I had a look at N3376 at 24.4.4 Iterator operations at page 815:

 template<class InputIterator> typename iterator_traits<InputIterator>::difference_type distance(InputIterator first, InputIterator last); 

Requires : If InputIterator meets the requirements of random access iterator, last shall be reachable from first or first shall be reachable from last ; otherwise, last shall be reachable from first .

Now I think that Requires is not fulfilled.

What does the standard state should happen in this case?

Answer 1

[iterator.requirements.general]:

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 .

The problem is that once you incremented v1.begin() v1.size()-1 times, the next increment operation induces undefined behavior, so v2.begin() cannot be reached from v1.begin() . The same argument makes v1.begin() unreachable from v2.begin() .

---

In case your question was "What happens if a condition in a Requires section is violated?", look at [res.on.required]:

Violation of the preconditions specified in a function's Requires: paragraph results in undefined behavior unless the function's Throws: paragraph specifies throwing an exception when the precondition is violated.

Answer 2

In some implementations of std::distance , the first iterator is incremented until it reaches the second iterator. The iterations are counted:

unsigned int counts = 0;
while (iter1 != iter2)
{
  ++counts;
  ++iter1;
}

If the iterators point to containers in different address spaces, the loop many not terminate. Using the terms in the standard, the second iterator is not reachable .

Answer 3

不满足requires ，这意味着代码具有未定义的行为：任何事情都可能发生。

Answer 4

In this case will be undefined behavior. Because last is not reachable from first by (possibly repeatedly) incrementing first .