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Is there a way to store a variadic template of a specific type C++(有没有办法存储特定类型的C++的可变模板)

转载 作者：bug小助手 更新时间：2023-10-26 21:13:52 24 4
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Lets say I have the following scenario where I am printing templates of type std::size_t.

假设我有以下场景，其中打印类型为std：：SIZE_t的模板。


template<std::size_t st>
void print_sizes(){
    std::cout << st << std::endl;
}

template<std::size_t st, std::size_t M, std::size_t... Rest>
void print_sizes(){
    std::cout << st << std::endl;
}

int main(){
    print_sizes<10, 12, 3>();
}

The above is entirely valid. However, I was wondering if there was a way to store the 10,12,3 variadic template like the following:

上述情况完全成立。然而，我想知道是否有一种方法来存储10，12，3可变模板，如下所示：


template<std::size_t st>
void print_sizes(){
    std::cout << st << std::endl;
}

template<std::size_t st, std::size_t M, std::size_t... Rest>
void print_sizes(){
    std::cout << st << std::endl;
    print_sizes<M, Rest...>();
}

using example = 10, 12, 3;
int main(){
    print_sizes<example>();
}

Obviously, that doesn't work and throws an error. Is there a way to do something like this though?

显然，这是不起作用的，并且会抛出错误。不过，有没有办法做到这一点呢？


更多回答

This is trivial. Wrap them into a type: template<std::size_t...> class wrapper {}; using example = wrapper<10, 12, 3>;.

这是微不足道的。将它们包装成一个类型：TEMPLATE类包装{}；Using Example=Wrapper<10，12，3>；。

优秀答案推荐

You can't easily store the 10, 12, 3 parameter pack itself, but you can store a template instantiated over those parameters:

您不能轻松地存储10、12、3参数包本身，但您可以存储在这些参数上实例化的模板：


#include <iostream>

template<std::size_t st>
void print_sizes(){std::cout<<st<<std::endl;}

template<std::size_t st, std::size_t M, std::size_t... Rest>
void print_sizes(){
    std::cout<<st<<std::endl;
    print_sizes<M, Rest...>();
}

int main(){
    auto f = print_sizes<10, 12, 3>; // save instantiated template

    f();                             // invoke it.
    return 0;
}

Note that if you can use C++17 or newer, you probably want to use a fold expression instead of the variadic template.

请注意，如果您可以使用C++17或更新版本，您可能希望使用fold表达式而不是可变参数模板。


Since you seem to want a compile-time list of things of one type, you might prefer an initializer list:

由于您似乎想要一种类型的东西的编译时列表，您可能更喜欢初始化列表：


#include <iostream>
#include <initializer_list>

void print_sizes(std::initializer_list<std::size_t> numbers) { 
    for (auto num : numbers)
        std::cout << num << "\n";
}

int main() { 
    print_sizes({10, 12, 3});
}


I think std::integer_sequence (since C++14) and fold expression (since C++17) might be what you are looking for:

我认为std：：INTEGER_SEQUENCE(从C++14开始)和Fold Expression(从C++17开始)可能就是您要寻找的：


Coliru

科里鲁


#include <iostream>

template <typename T, T... ints>
void print_sequence(std::integer_sequence<T, ints...> seq) {
  ((std::cout << ints << std::endl), ...);
}

constexpr auto seq = std::integer_sequence<size_t, 10, 12, 3>{};

int main() { print_sequence(seq); }

更多回答

These are good solutions, but the question is tagged C++11, so the answer should at least contain an alternative that works in C++11 as well.

这些都是很好的解决方案，但问题被标记为C++11，所以答案至少应该包含一个也适用于C++11的替代方案。

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