finish hw03 #24
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finish hw03 #24
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archibate
approved these changes
Jan 17, 2022
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| if constexpr (ops == Operator::Add) return a + b; | ||
| else if constexpr (ops == Operator::Sub) return a - b; | ||
| else if constexpr (ops == Operator::Mul) return a * b; | ||
| else if constexpr (ops == Operator::Div) return a / b; // RVO |
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这里其实可以声明 class Ops，然后传入 std::plus，std::minus，并调用 Ops{}(a + b)？
| auto x = a.cbegin(); | ||
| auto y = b.cbegin(); | ||
| for (; x != a.cend() && y != b.cend(); ++x, ++y) | ||
| out.push_back(std::move(cal_ops<T1, T2, ops>(*x, *y))); // NO RVO |
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应该不需要哈，在push_back参数中使用函数返回临时变量，会优化成调用push_back(&&)
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| template<class O, class V, class R, Operator ops, size_t...N> | ||
| constexpr decltype(auto) unpack_n_vector(O& o, V const& a, R const& b, std::index_sequence<N...>) noexcept { | ||
| static_cast<void>(std::initializer_list<int>{(op_vector_v<O, V, R, ops, N>(o, a, b), 0)...}); |
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使用 index_sequence 和 ... 语法实现的任意大小 variant 的匹配，没有用 std::visit，太神奇了！
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其实index_sequence 更适合序列中每个元素都存在，像variant这样的只有一个元素的用其他方法好一点。
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| if (N == a.index()) { | ||
| auto& e = std::get<N>(a); |
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不过我记得 std::visit 的实现用了函数指针什么的，总之是 O(1) 的，这样相当于每个都手动比较了一次是不是变成 O(n) 了？
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嗯，你说的对哈，测试了一下是变成O(n)了。通过index_sequence展开了N次，就要比较N次。我再想想有没有规避的方法哈
| } | ||
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| template <class... Args, class T> | ||
| constexpr decltype(auto) operator+(std::variant<Args...> const& a, std::vector<T> const& b) noexcept { |
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variant + vector？我觉得应该 variant<Args...> + T 然后约束 T in Args 比较好，这样 variant 里可以不是 vector。
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小彭老师，请检查hw03作业哈