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Copy vs move in std::pair braced initialization

April 11, 2020 by

Questions › Copy vs move in std::pair braced initialization
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Garmaine asked 3 years ago

I have the following code and its output printed below. I can't seem to understand why one set of braced initialization results in the move constructor being called, while the other results in the copy constructor. I have somewhat narrowed it down to direct-list-initialization vs copy-list-initialization per https://en.cppreference.com/w/cpp/language/list_initialization I just can't quite figure out which case my code belongs to. Thanks in advance.

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

struct Foo {
  Foo() {
    cout << "create foo\n";
  }

  ~Foo() {
    cout << "delete foo\n";
  }

  Foo(const Foo& f) {
    cout << "copy foo\n";
  }

  Foo(Foo&& f) noexcept {
    cout << "move foo\n";
  }

  Foo& operator=(const Foo& f) = delete;

  Foo& operator=(Foo&& f) = delete;
};

int32_t main() {
  pair<uint32_t, Foo> f1{0, Foo{}};  // Calls move ctor
  cout << "------------------------\n";

  pair<uint32_t, Foo> f2{0, {}};     // Calls copy ctor
  cout << "------------------------\n";

  return 0;
}

This results in

create foo

move foo

delete foo

------------------------

create foo

copy foo

delete foo

------------------------

delete foo

delete foo
Are you looking for the answer?
Original Question and Possible Answers can be found on `http://stackoverflow.com`

Question Tags: c++, initialization, move-semantics, std-pair, templates

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