我简化了我的代码产生错误,发现即使这个简单的计数函数也给我一个错误(见下文):
#include <boost/hana/tuple.hpp>
#include <boost/hana/fold.hpp>
#include <boost/hana/plus.hpp>
#include <boost/hana/integral_constant.hpp>
int main() {
using namespace boost;
constexpr auto inc = [](auto n, auto el) { return hana::int_c<n> + hana::int_c<1>; };
constexpr auto count = hana::fold(hana::make_tuple(hana::int_c<3>),
hana::int_<0>{},
inc
);
return 0;
}
错误(忽略了一些似乎无关紧要的事情):
/usr/local/include/boost/hana/detail/variadic/foldl1.hpp:202:57: error: ‘static constexpr decltype(auto) boost::hana::detail::variadic::foldl1_impl<2u>::apply(F&&, X1&&, X2&&) [with F = const main()::<lambda(auto:1, auto:2)>&; X1 = boost::hana::integral_constant<int, 0>; X2 = boost::hana::integral_constant<int, 3>]’ called in a constant expression
return foldl1_impl<sizeof...(xn) + 1>::apply(
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~^
static_cast<F&&>(f), static_cast<X1&&>(x1), static_cast<Xn&&>(xn)...
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
);
~
In file included from /usr/local/include/boost/hana/fold_left.hpp:18:0,
from /usr/local/include/boost/hana/concept/foldable.hpp:19,
from /usr/local/include/boost/hana/core/to.hpp:16,
from /usr/local/include/boost/hana/bool.hpp:17,
from /usr/local/include/boost/hana/tuple.hpp:16,
from ...:
/usr/local/include/boost/hana/detail/variadic/foldl1.hpp:31:41: note: ‘static constexpr decltype(auto) boost::hana::detail::variadic::foldl1_impl<2u>::apply(F&&, X1&&, X2&&) [with F = const main()::<lambda(auto:1, auto:2)>&; X1 = boost::hana::integral_constant<int, 0>; X2 = boost::hana::integral_constant<int, 3>]’ is not usable as a constexpr function because:
static constexpr decltype(auto) apply(F&& f, X1&& x1, X2&& x2) {
^~~~~
/usr/local/include/boost/hana/detail/variadic/foldl1.hpp:32:39: error: call to non-constexpr function ‘main()::<lambda(auto:1, auto:2)> [with auto:1 = boost::hana::integral_constant<int, 0>; auto:2 = boost::hana::integral_constant<int, 3>]’
return static_cast<F&&>(f)(static_cast<X1&&>(x1),
~~~~~~~~~~~~~~~~~~~^~~~~~~~~~~~~~~~~~~~~~~
static_cast<X2&&>(x2));
我使用c ++ 17和g ++ 6.3。看起来它说lambda并不被视为constexpr,但它看起来像它只使用常量值。任何人都可以告诉我如何使这段代码工作? (它的目的是计算传递给折叠的元组中的元素数量)
答案 0 :(得分:0)
如果你的编译器不支持constexpr lambdas,你必须自己写出闭包类型(在命名空间或类范围,而不是在函数范围,因为本地类不能有成员模板):
constexpr struct inc_t {
template<class N, class T>
constexpr auto operator()(N n, T) const { return hana::int_c<n> + hana::int_c<1>; }
} inc;
否则,您仍然可以使用hana,但计算结果将无法用于类型系统。