有没有办法帮助编译器处理复杂的参数推导？ [复制]

Question

我正在尝试构建编译时间表。我使用数组数组的数组......来做到这一点。不幸的是，编译器无法推断出结果类型的模板参数。有解决办法吗？

#include <array>

template <typename T, size_t... dims>
struct variadic_array;

template <typename T, size_t... dims>
using variadic_array_t = typename variadic_array<T, dims...>::type;

template <typename T, size_t dim>
struct variadic_array<T, dim> {
  using type = std::array<T, dim>;
};

template <typename T, size_t dim, size_t ...dims>
struct variadic_array<T, dim, dims...> {
  using type = std::array<variadic_array_t<T, dims...>, dim>;
};

template <typename T, size_t ...dims>
void foo(variadic_array_t<T, dims...>) {}

void call_foo() {
  foo(variadic_array_t<int, 3, 4>{});
}

编译错误：

error: no matching function for call to 'foo'
  foo(variadic_array_t<int, 3, 4>{});
  ^~~
note: candidate template ignored: couldn't infer template argument 'T'
  void foo(variadic_array_t<T, dims...>) {}
     ^

Answer 1

T 不能从 C<T>::type 推导出来（非推导上下文）。

您可以通过传递简单的模板T 并从中检索信息来解决问题：

// Helpers
template <typename T> struct is_std_array : std::false_type {};

template <typename T, std::size_t N>
struct is_std_array<std::array<T, N>> : std::true_type {};

template <typename T> struct identity { using type = T; };

// The trait
template <typename T, std::size_t...> struct variadic_array_traits;

template <typename T, std::size_t N, std::size_t ... dims>
struct variadic_array_traits<std::array<T, N>, dims...>
{
    using type =
        typename std::conditional_t<is_std_array<T>::value,
                                    variadic_array_traits<T, dims..., N>,
                                    identity<variadic_array<T, dims..., N>>>::type;    
};

// For SFINAE    
template <typename ... >
using void_t = void;

// Your method
template <typename T>
void_t<typename variadic_array_traits<T>::type>
foo(T)
{
    using variadic_array_type = typename variadic_array_traits<T>::type;
    // ...
}

Demo