C ++ 11在运行时索引元组而不使用开关的方法

2022-01-19 00:00:00 tuples c++ c++11

我有一段类似如下的 c++11 代码:

I have a piece of c++11 code similar like below:

switch(var) {
   case 1: dosomething(std::get<1>(tuple));
   case 2: dosomething(std::get<2>(tuple));
   ...
}

有什么办法可以去掉这个大开关吗?请注意，get<var> 不起作用，因为 var 不是常量，但我知道 var 的范围很小，即 (0-20).

Is there any way to remove this large switch ? Note that get<var> does not work because var is not constant, but I know var is in small range i.e. (0-20).

注意这里的重点是避免使用导致数组查找的数组...

Note that the point here is to avoid using an array that causes an array lookup...

关于性能的问题，有讨论函数数组在 if 和 switch 语句中的性能

well on the issue of performance, there is a discussion Performance of array of functions over if and switch statements

出于我自己的目的，我不争论哪个更好.

For my own purpose, I do not argue which one is better.

推荐答案

这是一个不使用索引序列的版本:

Here's a version that doesn't use an index sequence:

template <size_t I>
struct visit_impl
{
    template <typename T, typename F>
    static void visit(T& tup, size_t idx, F fun)
    {
        if (idx == I - 1) fun(std::get<I - 1>(tup));
        else visit_impl<I - 1>::visit(tup, idx, fun);
    }
};

template <>
struct visit_impl<0>
{
    template <typename T, typename F>
    static void visit(T& tup, size_t idx, F fun) { assert(false); }
};

template <typename F, typename... Ts>
void visit_at(std::tuple<Ts...> const& tup, size_t idx, F fun)
{
    visit_impl<sizeof...(Ts)>::visit(tup, idx, fun);
}

template <typename F, typename... Ts>
void visit_at(std::tuple<Ts...>& tup, size_t idx, F fun)
{
    visit_impl<sizeof...(Ts)>::visit(tup, idx, fun);
}

演示

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