C ++ 11方法在运行时索引元组，而不使用开关 [英] C++11 way to index tuple at runtime without using switch

问题描述

我有一块类似如下的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));
   ...
}

有任何方法来删除此大型开关？请注意， get< var> 不起作用，因为var不是常数，但我知道var在小范围ie（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).

请注意，这里的要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...

编辑：

在性能问题上，有一个讨论
函数数组在if和switch语句上的性能

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

为了我自己的目的，我不争论

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

推荐答案

这里有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);
}

DEMO

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