How to query a constexpr std::tuple at compile time?

不想你离开。 提交于 2019-12-18 11:46:15

问题


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


回答1:


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.




回答2:


I have not yet worked with C++0x, but it seems to me that std::get() is a function, rather than expression the compiler can directly interpret. As such, it has no meaning except at runtime, after the function itself has been compiled.



来源:https://stackoverflow.com/questions/5087364/how-to-query-a-constexpr-stdtuple-at-compile-time

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