The Standard says that std::tuple
has the following member functions
constexpr tuple();
explicit tuple(const Types&...);
Can someone please explain what is supposed to happen for std::tuple<>
?
A default constructor is a constructor that either has no parameters, or if it has parameters, all the parameters have default values. If no user-defined constructor exists for a class A and one is needed, the compiler implicitly declares a default parameterless constructor A::A() .
Class template std::tuple is a fixed-size collection of heterogeneous values. It is a generalization of std::pair. If std::is_trivially_destructible<Ti>::value is true for every Ti in Types , the destructor of tuple is trivial.
A default constructor is a constructor which can be called with no arguments (either defined with an empty parameter list, or with default arguments provided for every parameter). A type with a public default constructor is DefaultConstructible.
The new std::array and std::tuple containers provide developers with additional ways to manage structured data efficiently.
I guess the definition given in the standard is supposed to be pseudocode. That is the case with many of the definitions in the standard; it contains several requirements that are given verbally, but are satisfiable only with tricks like enable_if
. This seems to be an example where the C++-like pseudocode notation can actually lead to illegal C++ when trying to instantiate such an empty tuple (or it might just be an omission).
Both stdlibc++ and libc++ have an explicit specialization for the zero-element tuple. For example, in stdlibc++:
// Explicit specialization, zero-element tuple.
template<>
class tuple<>
{
public:
void swap(tuple&) noexcept { /* no-op */ }
};
with an implicitly-defined unambiguous default constructor.
Libc++ does not explicitly declare the parameterless default constructor. Presumably the templated constructor is then chosen as default constructor for non-empty tuples.
Interestingly, the two libraries disagree on what members the empty tuple has. For example, the following compiles with libc++, but not with libstdc++:
#include <tuple>
#include <memory>
int main() {
std::tuple<> t(std::allocator_arg, std::allocator<int>());
}
I believe this is a minor error in the standard. Clearly, when the Types
parameter pack is empty, the two constructor calls are equivalent and cannot be overloaded (see C++11 section 13). (Further note that the constructor using Types
is not a member template either --if it was, then it would be a legal overload.).
In other words, this code will not compile:
template <typename... Types>
struct Test
{
constexpr Test() {}
explicit Test(Types const&...) { /* etc. */ }
};
int main()
{
Test<> a;
Test<int> b;
}
e.g., a g++ v4.8 snapshot outputs:
tt.cxx: In instantiation of ‘struct Test<>’:
tt.cxx:10:10: required from here
tt.cxx:5:12: error: ‘Test<Types>::Test(const Types& ...) [with Types = {}]’ cannot be overloaded
explicit Test(Types const&...) { /* etc. */ }
^
tt.cxx:4:13: error: with ‘constexpr Test<Types>::Test() [with Types = {}]’
constexpr Test() {}
^
This can be fixed by using partial specialization:
template <typename... Types>
struct Test
{
constexpr Test() {} // default construct all elements
explicit Test(Types const&...) { /* etc. */ }
// and all other member definitions
};
template <>
struct Test<>
{
constexpr Test() {}
// and any other member definitions that make sense with no types
};
int main()
{
Test<> a;
Test<int> b;
}
which will compile correctly.
It appears the standard wanted a constexpr
default constructor was so that std::tuple<> var;
could be written instead of writing std::tuple<> var();
or std::tuple<> var{};
because of the use of explicit
with the other constructor. Unfortunately, its definition of std::tuple
does not work for tuples of size zero. The standard does permit such in section 20.4.2.7 (relational operators) though, "For any two zero-length tuples, [...]". Oops! :-)
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