c++ - 通过示例了解变量模板

Question

我试图通过以下示例了解变量模板的工作原理:

#include <iostream>

template <class T, const T& t>
int var = t.a;

struct T
{
    int a;
    constexpr T(): a(31){ }
};

T y;

const T t = y;

const T tt = T();

int main()
{ 
    std::cout << "var <T, t> = " << var<T, t> << std::endl;  //0
    std::cout << "y.a = " << y.a << std::endl;  //31
    std::cout <<"var <T, tt> = " << var<T, tt> << std::endl; //31
}

DEMO

老实说，我真的不知道这种行为。让我感到困惑的是特化 var<T, t>为 0，但 y.a是 31 .另外，如果我们初始化 T 类型的对象与临时我们也有不同的结果。你能澄清一下吗？

我的意思是，我正在寻找工作草案中的规范引用 N4296 ，描述该行为。

Answer 1

目前，变量模板的指定程度相当低。如果我们通过 the current core issues list ，我们看到了

• It is unclear whether and how a variable template can be defined multiple times in a program.
• It is unclear how definitions and declarations of variable templates are matched.
• It is unclear how variable template partial specializations are supposed to work.
• It is unclear what the point of instantiation of a variable template specialization is.

过去也不清楚初始化顺序变量模板遵循什么。 CWG issue 1744修改 [basic.start.init]/p2 以澄清这一点

Dynamic initialization of a non-local variable with static storage duration is unordered if the variable is an implicitly or explicitly instantiated specialization, and otherwise is ordered [Note: an explicitly specialized static data member or variable template specialization has ordered initialization. —end note].

var<T, t>是具有静态存储持续时间的非局部变量，是隐式实例化的特化。因此它的动态初始化是无序的。从 t不符合常量初始化的条件，这意味着 var<T, t>可能在t的动态初始化之前初始化, 结果为 0，无论 var 之间的相对顺序如何的定义和t的定义，并且无论 var<T, t> 的实例化点如何.

因此，moving the definition of var below the definition of t 和/或 an explicit instantiation of var<T, t> 对正在打印的内容没有影响，而 providing an explicit specialization for var<T, t> still initializing it to t.a 导致第一行打印 31 .