Why move constructor is generated when user-declared destructor is present?

Question

Following this answer and this move-constructor specification, there should be no implicit move constructor or move assignment operator.

However, the following code still compiles in gcc 7.2.1:

#include <vector>
#include <iostream>
using namespace std;

struct NoCopyNoMove
{
    NoCopyNoMove(const NoCopyNoMove&) = delete;
    NoCopyNoMove& operator=(const NoCopyNoMove&) = delete;
    NoCopyNoMove(NoCopyNoMove&&) = delete;
    NoCopyNoMove& operator=(NoCopyNoMove&&) = delete;

    NoCopyNoMove(int){};
};

struct NoCopy
{
    NoCopy(const NoCopyNoMove&) = delete;
    NoCopy& operator=(const NoCopyNoMove&) = delete;
    // NoCopy(NoCopy&&) = delete;
    // NoCopy& operator=(NoCopy&&) = delete;
    ~NoCopy() {
        std::cout << "decontructor" << std::endl;
    }
    NoCopy(int){};
};

int main()
{
    // vector<NoCopyNoMove> y; // fails!
    vector<NoCopy> y;
    y.emplace_back(1);
    y.emplace_back(2);
}

Why shouldn't it compile? `emplace_back` neither moves nor copies any `NoCopy` objects. By the way, it's called *destructor*, not *de**con**structor* — Fureeish, Oct 03 '18 at 22:02
@Fureeish an emplace_back could cause a reallocation which would require either move or copy of the existing elements. — Richard Hodges, Oct 03 '18 at 22:04

score 5 · Answer 1 · answered Oct 03 '18 at 22:12

You have a typo in your code. Where you have:

NoCopy(const NoCopyNoMove&) = delete;
NoCopy& operator=(const NoCopyNoMove&) = delete;

You should instead have:

NoCopy(const NoCopy&) = delete;
NoCopy& operator=(const NoCopy&) = delete;

Make those changes and the code indeed fails to compile:

https://wandbox.org/permlink/lVhPkL6ioLTC9k1w

Why move constructor is generated when user-declared destructor is present?

1 Answers1