Are copy constructors defined implicitly always, or only when they are used?

  • A+

Consider the following code:

#include <memory> #include <vector>  class A { private:   std::vector<std::unique_ptr<int>> _vals; };  int main() {   A a;   //A a2(a);   return 0; } 

Compiler A compiles this without issue unless I uncomment out the line A a2(a); at which point it complains about the copy constructor for std::unique_ptr being deleted, and therefore I can't copy construct A. Compiler B, however, makes that complaint even if I leave that line commented out. That is, compiler A only generates an implicitly defined copy constructor when I actually try to use it, whereas compiler B does so unconditionally. Which one is correct? Note that if I were to have used std::unique_ptr<int> _vals; instead of std::vector<std::unique_ptr<int>> _vals; both compilers correctly implicitly delete both copy constructor and assignment operator (std::unique_ptr has a explicitly deleted copy constructor, while std::vector does not).

(Note: Getting the code to compile in compiler B is easy enough - just explicitly delete the copy constructor and assignment operator, and it works correctly. That isn't the point of the question; it is to understand the correct behavior.)


From [class.copy.ctor]/12:

A copy/move constructor that is defaulted and not defined as deleted is implicitly defined when it is odr-used ([basic.def.odr]), when it is needed for constant evaluation ([expr.const]), or when it is explicitly defaulted after its first declaration.

A's copy constructor is defaulted, so it's implicitly defined only when it is odr-used. A a2(a); is just such an odr-use - so it's that statement that would trigger its definition, that would make the program ill-formed. Until the copy constructor is odr-used, it should not be defined.

Compiler B is wrong to reject the program.


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