hana简单折叠“调用非constexpr函数”

Question

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（省略了一些似乎无关紧要的）：

 /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));

我用c ++ 17和g ++ 6.3。 看起来它说lambda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）

Answer 1

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。