CODEWITHSUNDEEP
c++c++11move-semantics
Luke Skywalker
Luke Skywalker

Reputation: 1520

C++ copy constructor needed although not used

Anybody knows why the compiler needs a copy constructor for Foo in this situation:

#include <iostream>
#include <list>

class Foo {
public:
  Foo() {}
  Foo(const Foo &&f) noexcept {}
  Foo(const Foo &f) = delete;
  ~Foo() {}
};

void passFoo2(const Foo&& f) {
  std::list<Foo> l;
  l.push_back(std::move(f));
}

int main(int argc, char **argv) {
  Foo f;
  passFoo2(std::move(f));
  return 0;
}

The compiler (g++) complains the copy constructor is deleted. But it should not need it in that case?

So what am I missing here?

x@ubuntu:/tmp/c++$ g++ stackoverflow.cxx -std=c++11
In file included from /usr/include/c++/4.9/list:63:0,
             from stackoverflow.cxx:2:
/usr/include/c++/4.9/bits/stl_list.h: In instantiation of std::_List_node<_Tp>::_List_node(_Args&& ...) [with _Args = {const Foo&}; _Tp = Foo]:
/usr/include/c++/4.9/ext/new_allocator.h:120:4:   required from void __gnu_cxx::new_allocator<_Tp>::construct(_Up*, _Args&& ...) [with _Up = std::_List_node<Foo>; _Args = {const Foo&}; _Tp = std::_List_node<Foo>]
/usr/include/c++/4.9/bits/stl_list.h:514:8:   required from std::list<_Tp, _Alloc>::_Node* std::list<_Tp, _Alloc>::_M_create_node(_Args&& ...) [with _Args = {const Foo&}; _Tp = Foo; _Alloc = std::allocator<Foo>; std::list<_Tp, _Alloc>::_Node = std::_List_node<Foo>]
/usr/include/c++/4.9/bits/stl_list.h:1688:63:   required from void std::list<_Tp, _Alloc>::_M_insert(std::list<_Tp, _Alloc>::iterator, _Args&& ...) [with _Args = {const Foo&}; _Tp = Foo; _Alloc = std::allocator<Foo>; std::list<_Tp, _Alloc>::iterator = std::_List_iterator<Foo>]
/usr/include/c++/4.9/bits/stl_list.h:1029:9:   required from void std::list<_Tp, _Alloc>::push_back(const value_type&) [with _Tp = Foo; _Alloc = std::allocator<Foo>; std::list<_Tp, _Alloc>::value_type = Foo]
stackoverflow.cxx:14:30:   required from here
/usr/include/c++/4.9/bits/stl_list.h:114:71: error: use of deleted function Foo::Foo(const Foo&)
  : __detail::_List_node_base(), _M_data(std::forward<_Args>(__args)...) 
                                                                   ^
stackoverflow.cxx:8:3: note: declared here
   Foo(const Foo &f) = delete;
   ^

Upvotes: 3

Views: 367

Answers (2)

ABu
ABu

Reputation: 12299

The push_back method has two overloads:

void push_back( const T& value );
void push_back( T&& value );

If you use the first one, T (Foo in your case) must be "CopyInsertable". If you use the second one, it must be "MoveInsertable".

Your passFoo2 function receives a const Foo&& reference, and since it is const-qualified (see note below), the first overload is the best match. So, that overload is called, which requires your class Foo to be copyable.

Note: Function std::move transform a named-object to make it anonymous, and so, suitable to be binded by a rvalue-reference, but it doesn't change const qualifications (it is a const Foo anonymous object after the movement). For this reason the first overload is called.

Upvotes: 2

Richard Hodges
Richard Hodges

Reputation: 69912

the reason is that push_back has an overload for Foo&& not const Foo&&.

const Foo&& is superfluous. It's a legal type but its existence is an anachronism.

this compiles:

#include <iostream>
#include <list>

class Foo {
public:
    Foo() {}
    Foo( Foo &&f) noexcept {}
    Foo(const Foo &f) = delete;
    ~Foo() {}
};

void passFoo2(Foo&& f) {
    std::list<Foo> list;
    list.push_back(std::move(f));
}

int main(int argc, char **argv) {
    Foo f;
    passFoo2(std::move(f));
    return 0;
}

Upvotes: 8

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