c++ - 将模板嵌套类的方法定义移到声明之外

Question

是否可以将方法定义移到声明之外？

template <typename T1>
class A
{
  template <bool T2>
  class B;

  template<>
  class B<true>
  {
    void f() { /* Do smg */ }
  }
  class B<false>
  {
    void f() { /* Do smg else */ }
  }
}

如果我尝试在类声明之外定义 f()，就像这样

template <typename T1>
template <>
void A<T1>::B<true>::f() { /* Do smg */ }

编译器给出错误 C3855:模板参数 T2 与声明不兼容。

Answer 1

您不能显式特化非特化类模板的类成员模板，来自 [temp.expl.spec]:

In an explicit specialization declaration for a member of a class template or a member template that appears in namespace scope, the member template and some of its enclosing class templates may remain unspecialized, except that the declaration shall not explicitly specialize a class member template if its enclosing class templates are not explicitly specialized as well.

即使 A 定义中的 B 的显式特化也是错误格式的。如果您需要做这样的事情，我根本不会为 B 使用成员类模板。