C++ 模板作为模板的参数

Question

我运行遇到下面的问题。我有以下模板 Number:

template<int n>
struct Number
{
    static const int value = n;
};

现在假设我想在编译时添加两个这样的数字。具体来说，我想让以下代码段起作用：

//The following code should display 7:
std::cout << Number_add< Number<3>, Number<4> >::value << std::endl;

我试过类似的方法，但我的编译器不喜欢它。

template<Number<int> n1, Number<int> n2>
struct Number_add
{
    static const int value = n1::value + n2::value;
}

实施Number_add的正确方法是什么？我认为这里可能需要模板模板参数，但我也无法让它工作。将不胜感激。

Answer 1

Number<int> 不能用作 non-type template parameter because User defined classes aren't one of the allowed types. The allowed types are (reproduced from cppreference.com):

std::nullptr_t (since C++11);

integral type;

lvalue reference type (to object or to function);

pointer type (to object or to function);

pointer to member type (to member object or to member function);

enumeration type.

您可以简单地按照评论n.m中的建议进行操作

template<typename n1, typename n2>
struct Number_add
{
    static const int value = n1::value + n2::value;
}

Answer 2

template<int A, int B>
constexpr Number<A+B>
operator+(Number<A>, Number<B> ){ return {}; }

template<class Lhs, class Rhs>
using Number_add=decltype(Lhs{}+Rhs{});

std::cout << Number_add< Number<3>, Number<4> >::value << std::end;

打印 7.

C++ 模板作为模板的参数

C++ Templates as arguments for templates

c++

templates

compile-time

template-templates