出现错误：ISO C++ 禁止声明没有类型的“类型名称”

Question

我有一个嵌套的类定义，并且在对指向它的指针应用强制转换时出错。以下程序编译时出现错误：

test.cpp: In member function ‘void* Achild<T>::test(void*)’:
test.cpp:24:31: error: ISO C++ forbids declaration of ‘type name’ with no type [-fpermissive]
       ShortName::ptr = (const ShortName::Ptr*)input;
                               ^~~~~~~~~
test.cpp:24:25: error: expected primary-expression before ‘const’
       ShortName::ptr = (const ShortName::Ptr*)input;
                         ^~~~~
test.cpp:24:25: error: expected ‘)’ before ‘const’
       ShortName::ptr = (const ShortName::Ptr*)input;
                        ~^~~~~
                         )
test.cpp:25:6: warning: no return statement in function returning non-void [-Wreturn-type]
      }

我不明白为什么我在第 24 行出现错误。任何帮助将不胜感激！

template<typename T>
   class VeryLongName
   {
     public:
         class Ptr
         {
                 public:
                         int a;
                         Ptr() = default;
         };
         const Ptr* ptr;

   };

   template <typename T>
   class Achild: public VeryLongName<T>
   {
     using ShortName = VeryLongName<T>;
   public:
     Achild<T>() = default;

     void test(void * input)
     {
             ShortName::ptr = (const ShortName::Ptr*)input;
     }

   };

int main()
{
        auto *achild = new Achild<int>();
        auto a = new VeryLongName<int>::Ptr();
        achild->test((void*)a);
}

Answer 1

您缺少typename声明：

ShortName::ptr = (const typename ShortName::Ptr*) input;

因为ShortName取决于你的模板类型。