unordered_set :: erase（pos）是否保留元素的顺序？

Question

I read in the standard of C++14 that the order of elements is preserved when using erase(iterator pos) of unordered_set .

I tried the following code with g++-6.2.0 and clang-3.9 (on linux though, this gcc's stdlib). Both should be able to handle that by the C++14-spec, I think:

#include <unordered_set>
#include <iostream>
using std::unordered_set; using std::cout;

// output
template<typename Elem, typename Comp>
std::ostream& operator<<(std::ostream&os, const unordered_set<Elem,Comp>&data) {
    for(auto &e : data) { os << e << ' '; } return os << '\n'; }

int main() {
  unordered_set<int> nums{ 1,2,3,4,5,6,7,8,9,10 };
  cout << nums; // MSVC: 9 1 2 3 4 5 6 7 8 10
  for(auto it = nums.begin(); it!=nums.end(); ++it) {
    if(*it % 2 == 0) {
      nums.erase(it);
    }
  }
  cout << nums; // MSCV: 9 1 3 5 7
}

Yes, the order of the elements is arbitrary. Here MSVC++ 19.00 had 9 1 2 3 4 5 6 7 8 10 . And after erasing all even elements the remaining elements are still in the same order 9 1 3 5 7 .

With g++ and clang++ though, I got a completely bad output of

10 9 8 7 6 5 4 3 2 1
9 8 7 6 5 4 3 2 1

which seems to indicate that the order of the elements was not preserved between calls but just... I don't know.

What is going on?

Answer 1

I suppose that this cycle is wrong:

for(auto it = nums.begin(); it!=nums.end(); ++it) {
    if(*it % 2 == 0) {
        nums.erase(it);
    }
}

If erase is performed then it is invalidated and you cannot increment it. Presumably it causes the aforementioned behaviour.

You should use something like this:

for(auto it = nums.begin(); it!=nums.end();) {
    if(*it % 2 == 0) {
        nums.erase(it++);
    } else {
        ++it;
    }
}