在我的项目中,有一个由模板定义的函数来查找元组的索引,我还是不明白它是如何工作的: 似乎有一个递归,但我不知道它是如何终止在正确的索引?
.schedule {
position: relative;
padding-bottom: 56.25%; /* 16:9 */
padding-top: 25px;
height: 0;
}
.schedule iframe {
position: absolute;
top: 0;
left: 0;
width: 100%;
height: 100%;
}
答案 0 :(得分:3)
专业化是终止条件。请注意,它要求First
等于Search
:
type_index<Index, Search, Search, Types ...>
^^^^^^ ^^^^^^
例如,如果您从
开始type_index<0, C, A, B, C, D>,
这不符合专业化,因此将使用通用模板,重定向(通过其type
成员)
type_index<0, C, A, B, C, D>::type = type_index<1, C, B, C, D>::type
但是直到链到达
才能评估type_index<0, C, A, B, C, D>::type = ... = type_index<2, C, C, D>::type
此时可以使用部分特化,即
type_index<2, C, C, D>::type = type_index<2, C, C, D>
type_index<2, C, C, D>::index = 2
等等
type_index<0, C, A, B, C, D>::type::index = 2
^ ^ ^ ^
0 1 2 3
按预期。
请注意,您无需携带Index
,确实可以删除整个::type
内容:
template<typename, typename...>
struct type_index;
template<typename Search, typename Head, typename... Tail>
struct type_index<Search, Head, Tail...> {
// Search ≠ Head: try with others, adding 1 to the result
static constexpr size_t index = 1 + type_index<Search, Tail...>::index;
};
template<typename Search, typename... Others>
struct type_index<Search, Search, Others...> {
// Search = Head: if we're called directly, the index is 0,
// otherwise the 1 + 1 + ... will do the trick
static constexpr size_t index = 0;
};
template<typename Search>
struct type_index<Search> {
// Not found: let the compiler conveniently say "there's no index".
};
这适用于:
type_index<C, A, B, C, D>::index
= 1 + type_index<C, B, C, D>::index
= 1 + 1 + type_index<C, C, D>::index
= 1 + 1 + 0
= 2
如果类型不在列表中,则会说(GCC 6.2.1):
In instantiation of ‘constexpr const size_t type_index<X, C>::index’:
recursively required from ‘constexpr const size_t type_index<X, B, C>::index’
required from ‘constexpr const size_t type_index<X, A, B, C>::index’
required from [somewhere]
error: ‘index’ is not a member of ‘type_index<X>’
我觉得很清楚。