我有一个在 typename ContainerType
上参数化的函数模板,它接受 const ContainerType&
参数。它将容器内的每个元素转换为其他类型,并(当前)返回转换后类型的 std::vector
。
有什么方法可以“借用”推导的 ContainerType 的实际外部类模板吗?
我目前有this code returning the hardcoded vector
:
template<typename ContainerType, typename OutputType = decltype(convert(std::declval<typename ContainerType::value_type>()))>
std::vector<OutputType> convert(const ContainerType& input)
{
std::vector<OutputType> result;
result.reserve(input.size());
std::transform(input.begin(), input.end(),
std::back_inserter(result),
static_cast<OutputType(*)(typename ContainerType::value_type)>(&convert));
return result;
}
我想推断出仅从其 value_type
中剥离的类模板。请记住,容器通常有超过 1 个模板参数(并且数量因容器和实现而异),因此 ContainerType
的模板参数不是此处的解决方案。
最佳答案
这简直太糟糕了,而且不适用于 map 。我建议重新考虑你的界面。
template<bool isAlloc, class From, class Of, class With>
struct rebind_arg { using type = From; };
template<bool isAlloc, class From, class Of, class With>
using rebind_arg_t = typename rebind_arg<isAlloc, From, Of, With>::type;
// allocators need to be rebound with allocator_traits
template<class From, class Of, class With>
struct rebind_arg<true, From, Of, With> {
using type = typename std::allocator_traits<From>::template rebind_alloc<With>;
};
// Try to rebind non-allocator arguments
template<class T, class With>
struct rebind_arg<false, T, T, With> { using type = With; };
template<template<class...> class X, class T, class... Args, class With>
struct rebind_arg<false, X<Args...>, T, With> {
using type = X<rebind_arg_t<false, Args, T, With>...>;
};
// resolve an ambiguity
template<template<class...> class X, class... Args, class With>
struct rebind_arg<false, X<Args...>, X<Args...>, With> {
using type = With;
};
// Obtain the container's allocator type if it has one
template<class T, class=void>
struct get_allocator_type { struct none; using type = none; };
template<class T>
struct get_allocator_type<T, std::void_t<typename T::allocator_type>> {
using type = typename T::allocator_type;
};
// Check if a type is the allocator type of another type
template<class T, class C>
constexpr bool is_allocator_of = std::is_same_v<T, typename get_allocator_type<C>::type>;
template<class C, class With>
struct rebind_container;
template<template<class...> class X, class... Args, class With>
struct rebind_container<X<Args...>, With> {
using Source = X<Args...>;
using type = X<rebind_arg_t<is_allocator_of<Args, Source>, Args,
typename Source::value_type, With>...>;
static_assert(!std::is_same_v<Source, type>, "Rebinding unsuccessful");
};
template<class C, class With>
using rebind_container_t = typename rebind_container<C, With>::type;
关于c++ - 有没有办法获取容器模板类型以将其与另一个 value_type 重用?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/48241349/