In C++0x, one can create a constexpr std::tuple, e.g. like
#include <tuple>
constexpr int i = 10;
constexpr float f = 2.4f;
constexpr double d = -10.4;
constexpr std::tuple<int, float, double> tup(i, f, d);
One also can query a std::tuple at runtime, e.g. via
int i2 = std::get<0>(tup);
But it is not possible to query it at compile time, e.g.,
constexpr int i2 = std::get<0>(tup);
will throw a compilation error (at least with the latest g++ snapshot 2011-02-19).
Is there any other way to query a constexpr std::tuple at compile time?
And if not, is there a conceptual reason why one is not supposed to query it?
(I am aware of avoiding using std::tuple, e.g., by using boost::mpl or boost::fusion instead, but somehow it sounds wrong not to use the tuple class in the new standard...).
By the way, does anybody know why
constexpr std::tuple<int, float, double> tup(i, f, d);
compiles fine, but
constexpr std::tuple<int, float, double> tup(10, 2.4f, -10.4);
not?
Thanks a lot in advance! - lars
std::get
is not marked constexpr
, so you cannot use it to retrieve the values from a tuple
in a constexpr
context, even if that tuple is itself constexpr
.
Unfortunately, the implementation of std::tuple
is opaque, so you cannot write your own accessors either.
Now, std::get<> is a constexpr function. The following code compiles for me if I use gcc c++ 11 or above.
constexpr int i2 = std::get<0>(tup);
constexpr std::tuple<int, float, double> tup(10, 2.4f, -10.4);
Furthermore, you can generate a list of numbers at compile time by using make_index_sequence (c++14 or above) and access the tuple.
constexpr auto size = std::tuple_size<decltype(tup)>::value;
for_sequence(std::make_index_sequence<size>{}, [&](auto i){
constexpr auto property = std::get<i>(tup);
std::cout<<property<<std::endl;
});
template <typename T, T... S, typename F>
constexpr void for_sequence(std::integer_sequence<T, S...>, F&& f) {
using unpack_t = int[];
(void)unpack_t{(static_cast<void>(f(std::integral_constant<T, S>{})), 0)..., 0};
}
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