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每种可变参数模板的调用权模板专门化

[英]Call right template specialization for each type of a variadic template

I have a function foo() that takes a list of types T... and inside calls another (templated) function called do_stuff() for every element of a vector that is passed in. More specifically, we loop over the vector (of length sizeof...(T) ), and would like to call do_stuff<Ti>() for vector[i] , where Ti is the i 'th type in T... 我有一个函数foo() ,它接受类型T...的列表,并针对传入的向量的每个元素调用另一个(模板化的)函数do_stuff() 。更具体地说,我们遍历向量(长度sizeof...(T) ),并想对vector[i]调用do_stuff<Ti>() ,其中TiT...的第i个类型T...

The information is available at compile time so I guess this is possible, but how we do it nicely? 该信息在编译时可用,所以我想这是可能的,但是我们如何做到这一点呢?

#include <iostream>
#include <string>
#include <vector>
#include <cassert>

template <typename T>
T do_stuff(int param);

template <>
int do_stuff(int param)
{
    return int(100);
}

template <>
std::string do_stuff(int param)
{
    return std::string("foo");
}

template <typename... T>
void foo(const std::vector<int>& p)
{
    assert(p.size() == sizeof...(T));
    for (int i = 0; i < p.size(); ++i)
    {
        // Won't compile as T is not specified:
        //do_stuff(p[i]);
        // How do we choose the right T, in this case Ti from T...?
    }
}

int main()
{
    std::vector<int> params = { 0,1,0,5 };
    foo<int, std::string, std::string, int>(params);
}

You can use a C++17 fold expression: 您可以使用C ++ 17折叠表达式:

template <typename... T>
void foo(const std::vector<int>& p)
{
    assert(p.size() == sizeof...(T));

    std::size_t i{};
    (do_stuff<T>(p[i++]), ...);
}

live example on godbolt.org Godbolt.org上的实时示例


Alternatively, you can avoid the mutable i variable with std::index_sequence : 另外,您可以使用std::index_sequence避免使用可变的i变量:

template <typename... T>
void foo(const std::vector<int>& p)
{
    assert(p.size() == sizeof...(T));

    [&p]<auto... Is>(std::index_sequence<Is...>)
    {
        (do_stuff<T>(p[Is]), ...);
    }(std::index_sequence_for<T...>{});
}

live example on godbolt.org Godbolt.org上的实时示例

What about as follows ? 如下呢?

template <typename ... T>
void foo (std::vector<int> const & p)
{
    assert(p.size() == sizeof...(T));

    using unused = int[];

    std::size_t  i{ 0u };

    (void)unused { 0, ((void)do_stuff<T>(p[i++]), 0)... };
}

If you can use C++17, see the Vittorio Romeo's answer for a more elegant and concise solution. 如果可以使用C ++ 17,请参见Vittorio Romeo的答案,以获取更优雅,更简洁的解决方案。

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