Returning temporaries of type with deleted move/copy ctor

Question

Consider the following program:

#include<iostream>
using namespace std;

struct S
{
    S() = default;
    S(const S& other) = delete;
    S(S&& other) = delete;
    int i;
};

S nakedBrace()
{
     return {}; // no S constructed here?
}

S typedBrace()
{
     return S{};
}

int main()
{
    // produce an observable effect.
    cout << nakedBrace().i << endl; // ok
    cout << typedBrace().i << endl; // error: deleted move ctor
}

Sample session:

$ g++ -Wall -std=c++14 -o no-copy-ctor no-copy-ctor.cpp
no-copy-ctor.cpp: In function 'S typedBrace()':
no-copy-ctor.cpp:19:12: error: use of deleted function 'S::S(S&&)'
   return S{};
            ^
no-copy-ctor.cpp:8:5: note: declared here
     S(S&& other) = delete;

It astonishes me that gcc accepts nakedBrace(). I thought that conceptually the two functions are equivalent: A temporary S is constructed and returned. Copy elision may or may not be performed, but the move or copy ctor (both are deleted here) must still be accessible, as mandated by the standard (12.8/32).

Does that mean that nakedBrace() never constructs an S? Or it does, but directly in the return value with brace-initialization, so that no copy move/ctor is conceptually required?

Answer 1

This is standard behaviour.

N4140 [stmt.return]/2: [...] A return statement with a braced-init-list initializes the object or reference to be returned from the function by copy-list-initialization (8.5.4) from the specified initializer list. [...]

This means that the initializations carried out by nakedBrace and typedBrace are equivalent to these:

S nakedBrace = {}; //calls default constructor
S typedBrace = S{}; //calls default, then copy constructor (likely elided)