具有非类型模板参数的构造函数 [英] Constructor with non-type template arguments

问题描述

在此问题中，它表明不可能仅将模板参数直接用于类构造函数，因为如果你写类似

In this question it's stated that it's impossible to just directly use template arguments for class constructor, because if you write something like

struct S{
    template<typename T>
    S() { ... }
}

然后，您将无法调用此构造函数.但是，有一些变通办法可以使此工作生效，例如，通过模板参数推导.

Then you have no way of calling this constructor. However, there're some workarounds to make this work, for example, through template argument deduction.

但是我所知道的所有这些变通方法仅适用于类型实参.所以，问题是

But all of these workarounds I know are for type arguments only. So, the question is

是否有任何变通办法可以使非类型模板参数起作用?

struct S{
    template<int x>
    S() { ... }
}

我对应该在现代C ++(C ++ 17标准，包括所有TS)中使用的解决方案感兴趣，因为这是一个理论问题，而不是实践问题.

I'm interested in solutions which should work in modern C++ (C++17 standard, including all TS), as this is a theoretical rather than practical question.

推荐答案

但是我知道的所有这些变通方法仅适用于类型参数

But all of these workarounds I know are for type arguments only

解决方法都不是特定于类型的-关键是在构造函数中粘贴一些可以推论的内容.因此，如果需要类型，可以执行以下操作:

None of the workarounds are type-specific - the point is to stick something in the constructor that can be deduced. So if we want a type, we do something like:

template <class T> struct tag { };

struct S {
    template <class T>
    S(tag<T>);
};

，如果我们想要一个 int ，我们可以做同样的事情:

and if we want an int, we do the same thing:

template <int I> struct val { };

struct S {
    template <int I>
    S(val<I>);
};

对于值，您甚至不需要自己的标记类型-您可以在 std :: integral_constant 的顶部打包.

For values, you don't even need to come up with your own tag type - you can piggy-pack on top of std::integral_constant.

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