Given the following code:
struct A;
struct B {
B() {}
B(A &&) {}
};
struct A {
A() {}
A(B &&) {}
};
Then I can use as many braces as I want to construct A or B.
// default construct A
auto a = A{};
// default construct B, forward to A
auto b = A{{}};
// default construct A, forward to B, forward to A
auto c = A{{{}}};
// etc.
auto d = A{{{{}}}};
auto e = A{{{{{}}}}};
Similarly, given
struct C {
C(std::initializer_list<C>) {}
};
then I can also use as many braces as I want
// default construct C
auto f = C{};
// construct from initializer_list of one default constructed C
auto g = C{{}};
// construct from initializer_list of one C constructed from empty initializer_list
auto h = C{{{}}};
// etc.
auto i = C{{{{}}}};
auto j = C{{{{{}}}}};
Why doesn't the same argument work for a truly boring type?
struct D {
};
or, rewritten for clarity:
struct D {
D() {}
D(D &&) {}
};
This fails even on
auto k = D{{}};
Why does this not default construct a D with the innermost braces, and then pass that rvalue on to the move constructor of D?
See it live: https://godbolt.org/z/E763EPGh1