std::ranges::concat_view<Views...>::begin
来自 cppreference.cn
< cpp | ranges | concat view
constexpr /*iterator*/<false> begin() requires (!(/*simple-view*/<Views> && ...)); |
(1) | (since C++26) |
constexpr /*iterator*/<true> begin() const requires (ranges::range<const Views> && ...) && |
(2) | (since C++26) |
返回指向 concat_view
开头的迭代器。
1) 等价于
it.template
return it;.
iterator
<false> it(this, std::in_place_index<0>,
ranges::begin(std::get<0>(views_
)));it.template
satisfy
<0>();return it;.
2) 等价于
it.template
return it;.
iterator
<true> it(this, std::in_place_index<0>,
ranges::begin(std::get<0>(views_
)));it.template
satisfy
<0>();return it;.
[编辑] 返回值
如上所述。
[编辑] 示例
初步版本可以在 Compiler Explorer 上查看。
运行此代码
#include <ranges> #include <string_view> using namespace std::literals; int main() { static constexpr auto c = {"🐱", "🐶"}; static constexpr auto a = {"🤡"sv}; static constexpr auto t = {"💚"sv}; static constexpr auto cat{std::views::concat(c, a, t)}; static_assert(*cat.begin() == "\U0001F431" and cat.begin()[1] == "🐶" and *(cat.begin() + 2) == "\N{CLOWN FACE}"); }
[编辑] 参见
返回指向末尾的迭代器或哨兵 (public member function) | |
(C++20) |
返回指向范围开头的迭代器 (customization point object) |