Explicit copy constructor and std::sort

Question

When sorting a container of objects having an explicit copy ctor I get compiler errors (from g++ 4.8.2 and clang++ 3.4, both in -std=c++11 mode) that I don't understand. I've created a simple example to demonstrate the problem

class A {
public:
  explicit A(int i): m_i(i) {};
  explicit A(const A& other): m_i(other.m_i) {};
  int i() const {return m_i;};
private:
  int m_i;
};

bool is_less(const A& a, const A& b) {
  return a.i() < b.i();
}

int main(int, char*[]) {
  std::vector<A> objects;
  objects.push_back(A(3));
  objects.push_back(A(5));
  objects.push_back(A(-1));

  std::cout << is_less(objects[1], objects[2]);
  std::sort(objects.begin(), objects.end(), is_less);

  for (auto& a: objects) {
    std::cout << a.i() << " ";
  }
  std::cout << std::endl;
}

This fails with

error: 
  no matching constructor for initialization of '_ValueType' (aka 'A')

in clang++ and with

error: no matching function for call to ‘A::A(std::remove_reference<A&>::type)

in g++. The code compiles and works fine if the copy constructor is not explicit (but I want to enforce that only references to my objects can be used as parameters and return values). The code also compiles after removing the call to std::sort (so is_less(objects[1], objects[2]) is not a problem). Hence my question is what std::sort does when calling the comparing function that makes compiling this code fail and how to fix it.

After a lot of research, the only question that came close to my problem is In copy-initialization, is the call to the copy constructor explicit or implicit? which links to a bug in gcc. However, clang shows the same behavior, so I'd really like to understand what's going on.

Only crazy people make copy constructors `explicit`, such types are not CopyConstructible — Jonathan Wakely, Aug 27 '14 at 14:38

interjay · Accepted Answer · 2014-08-27T14:51:39.247

12

std::sort requires that the element type be MoveConstructible.

The requirements for MoveConstructible state that the expression T u = rv; must be valid. However, this expression performs copy initialization and requires that there is a non-explicit copy or move constructor.

In this case, the copy constructor is explicit, and declaring it means that there is no implicitly-declared move constructor. Therefore, the expression is invalid, and the class A is not MoveConstructible.

edited Aug 27 '14 at 14:51

answered Aug 27 '14 at 14:39

interjay

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7

What makes the expression invalid is not just that the copy constructor is explicit, but also that as a side effect of merely having a copy constructor in the first place, the move constructor is disabled. With a move constructor and move assignment operator, it does work, even if the copy constructor is still explicit. – Aug 27 '14 at 14:42

Explicit copy constructor and std::sort

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