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https://github.com/changkun/modern-cpp-tutorial.git
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update: move std::variant to container chapter
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@@ -228,22 +228,27 @@ int index = 1;
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std::get<index>(t);
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```
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那么要怎么处理?答案是,**标准库做不到**。这里介绍一个使用 `boost::variant` 配合变长模板参数的黑魔法:
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那么要怎么处理?答案是,使用 `std::variant<>`(C++ 17 引入),提供给 `variant<>` 的类型模板参数
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可以让一个 `variant<>` 从而容纳提供的几种类型的变量(在其他语言,例如 Python/JavaScript 等,表现为动态类型):
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```cpp
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#include <boost/variant.hpp>
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#include <variant>
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template <size_t n, typename... T>
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boost::variant<T...> _tuple_index(size_t i, const std::tuple<T...>& tpl) {
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if (i == n)
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return std::get<n>(tpl);
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else if (n == sizeof...(T) - 1)
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throw std::out_of_range("越界.");
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else
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return _tuple_index<(n < sizeof...(T)-1 ? n+1 : 0)>(i, tpl);
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constexpr std::variant<T...> _tuple_index(const std::tuple<T...>& tpl, size_t i) {
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if constexpr (n >= sizeof...(T))
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throw std::out_of_range("越界.");
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if (i == n)
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return std::variant<T...>{ std::in_place_index<n>, std::get<n>(tpl) };
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return _tuple_index<(n < sizeof...(T)-1 ? n+1 : 0)>(tpl, i);
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}
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template <typename... T>
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boost::variant<T...> tuple_index(size_t i, const std::tuple<T...>& tpl) {
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return _tuple_index<0>(i, tpl);
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constexpr std::variant<T...> tuple_index(const std::tuple<T...>& tpl, size_t i) {
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return _tuple_index<0>(tpl, i);
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}
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template <typename T0, typename ... Ts>
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std::ostream & operator<< (std::ostream & s, std::variant<T0, Ts...> const & v) {
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std::visit([&](auto && x){ s << x;}, v);
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return s;
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}
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```
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@@ -251,7 +256,7 @@ boost::variant<T...> tuple_index(size_t i, const std::tuple<T...>& tpl) {
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```cpp
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int i = 1;
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std::cout << tuple_index(i, t) << std::endl;
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std::cout << tuple_index(t, i) << std::endl;
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```
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### 元组合并与遍历
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