采用地址时，模板类型（类/函数）的实例化规则是什么？

问题描述

在回答问题时，我遇到了这个问题

In answering this question, I came across this difference in behaviour with respect to template instantiation.

最初有一个函数模板

template <typename T> void my_callback(void* data) { … }

现在有些东西需要这个地址 - code> void * ，所以显而易见的方法是

Now something requires the address of this - specifically a void*, so the obvious approach is

bar(reinterpret_cast<void*>(&my_callback<int>));

但是，对于编译器版本pre gcc 4.5，这将失败与一个不足够的上下文...错误。好 - 修正是强制第一个 - 强制实例化，即：

However, with compiler versions pre gcc 4.5, this fails with a not-enough context... error. Fine - so the fix is to "cast" first - which forces instantiation, i.e:

void (*callback)(void*) = my_callback<int>;
bar(reinterpret_cast<void*>(callback));

这很好。

现在第二种情况，而不是一个自由函数，它是一个类模板的静态成员，即

Now the second scenario, rather than being a free function, it's a static member of a class template, i.e.

template <typename T>
struct foo
{
  static void my_callback(void* data) {
    T& x = *static_cast<T*>(data);
    std:: cout << "Call[T] with " << x << std::endl;
  }
};

现在，原来的 reinterpret_cast

bar(reinterpret_cast<void*>(&foo<int>::my_callback));

所以我的问题是 - 为什么这种明显的行为差异？

So my question is - why this apparent difference in behaviour?

推荐答案

从n3290,14.7.1隐式实例化[temp.inst]

From n3290, 14.7.1 Implicit instantiation [temp.inst]

在第1段中有类模板特化的类似规则。注意，该标准以专业化的方式表示，因为当使用模板时隐含地声明专业化，至少对于功能模板（第8段）没有用户提供的专门化。

There are similar rules in paragraph 1 for class template specializations. Notice that the Standard speaks in terms of specialization because a specialization is implicitly declared when a template is used an no user-provided specialization is here, at least for function templates (paragraph 8).

结合第10段，

我认为经验法则是：只要一个对象/类成员/函数需要或者使程序以其他方式工作（非正式地讲话），模板就被隐式实例化， 。

I think the rule of thumb is: as soon as an object/class member/function is needed or to make the program otherwise work (speaking informally), the template is implicitly instantiated but no sooner. This include taking the address of a function.

对于你链接的问题， reinterpret_cast 的一些用法可能会使程序不符合，此时与提示实例无关 - / shameless>。

As to the question you linked, some uses of reinterpret_cast may make the program non-conformant, by which time it's irrelevant to mention instantiations -- I invite you to see my answer there</shameless>.

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