On 8 February 2016 at 22:40, H.J. Lu <hjl.to...@gmail.com> wrote:
> "empty type". An empty type is either an array of empty types or a
> class type where every member is of empty type.
Note that the term "empty type" is commonly used in type theory to
denote a (or the) type with no values. The closest thing C has would be
an empty enum when using -fstrict-enums. (Declaring it as return type
implies [[noreturn]] or undefined behaviour.)
A type with a unique value (such as void or an empty struct) is usually
known as a unit type.
BTW, being standard layout is not sufficient (nor required afaict) for
zero-register passing of a unit type. The requirement you need is
trivially-copyable. Example:
#include <map>
#include <iostream>
#include <type_traits>
using namespace std;
class EmptyInt {
static map< const EmptyInt *, int > values;
public:
EmptyInt() = default;
EmptyInt( int x ) { values[this] = x; }
~EmptyInt() { values.erase(this); }
operator int () const { return values[this]; }
};
typeof( EmptyInt::values ) EmptyInt::values;
EmptyInt foo() {
return 42;
}
int main() {
cout << is_standard_layout<EmptyInt>{} << endl;
cout << foo() << endl;
return 0;
}
This evil contraption satisfies all POD-requirements except for not
being trivially-copyable. On the other hand taking this example from
http://en.cppreference.com/w/cpp/concept/StandardLayoutType
struct Q {};
struct S : Q {};
struct T : Q {};
struct U : S, T {}; // not a standard-layout class
Even though U is not standard-layout, it is trivially-copyable and I see
no reason to allocate a register to pass it.
Matthijs van Duin
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