Why is the move constructor defined and the assignment operator implicitly deleted?

Question

struct Foo { Foo() = default; Foo(Foo&&) = default; }; int main() { Foo a, b; a = b; // ^ return 0; }

error: use of deleted function 'Foo& Foo::operator=(const Foo&)'

in g++4.6 -std=c++0x it's ok. but, in g++6.2 -std=c++11 it's error. why?

Read further : "note: ‘constexpr Foo& Foo::operator=(const Foo&)’ is implicitly declared as deleted because ‘Foo’ declares a move constructor or move assignment operator". And that is because of the standard says so : stackoverflow.com/questions/11255027/… — Hatted Rooster
– Hatted Rooster, Commented May 13, 2017 at 15:28

Sam Varshavchik · Accepted Answer · 2017-05-13 15:32:14Z

The answer is because the C++ standard says so:

[class.copy]

If the class definition does not explicitly declare a copy constructor, one is declared implicitly. If the class definition declares a move constructor or move assignment operator, the implicitly declared copy constructor is defined as deleted; otherwise, it is defined as defaulted (8.4).

You can always declare a default copy constructor, in your case:

 Foo(const Foo&) = default;

Lightness Races in Orbit · Accepted Answer · 2017-05-13 15:36:38Z

Sam has explained why GCC 6.2's behaviour is correct.

GCC 4.6's behaviour is simply due to incomplete C++11 support in that version (as evidenced by the "experimental" C++0x switch); C++11 was not fully supported until 4.8.

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