C++模板专业化:编译错误:"不是类型"

Question

如果我删除模板特化部分(试图打印"测试2"的部分),代码编译很好,但我希望能够有一个特殊情况,运行一个看起来干净的外部用户不同的代码路径.

#include <iostream>

using namespace std;

struct SpecialType {};

template<typename A , typename B = SpecialType>
class Test
{
public:
    class TestInner
    {
    public:
        TestInner& operator* ();
    };
};

template<typename A , typename B>
typename Test<A , B>::TestInner& Test<A , B>::TestInner::operator* ()
{
    cout << "Test 1" << endl;
    return *this;
}

// If the following is removed, everything compiles/works, but I want this alternate code path:
template<typename A>
typename Test<A , SpecialType>::TestInner& Test<A , SpecialType>::TestInner::operator* ()
{
    cout << "Test 2" << endl;
    return *this;
}

int main()
{
    Test<int , SpecialType>::TestInner test;
    *test;

   return 0;
}

我究竟做错了什么？

编辑:顺便说一下,编译器错误读取:

main.cpp:26:44: error: 'Test<A, SpecialType>::TestInner' is not a type
 typename Test<A , SpecialType>::TestInner& Test<A , SpecialType>::TestInner::operator* ()
                                            ^
main.cpp:26:89: error: invalid use of dependent type 'typename Test<A, SpecialType>::TestInner'
 typename Test<A , SpecialType>::TestInner& Test<A , SpecialType>::TestInner::operator* ()
                                                                                         ^

Answer 1

为专门类添加声明:

template<typename A>
class Test<A, SpecialType>
{
public:
    class TestInner
    {
    public:
        TestInner& operator* ();
    };
};

问题是您为未声明的特化定义了一个成员.模板化类的特化不会与通用模板共享任何成员或方法,因此通用模板的声明不会作为该模板类的任何特化的声明.

考虑一下:

template <class T>
class Foo {
  void GeneralFunction(T x);
}

和专业化:

template <>
class Foo<int> {
  void SpecialisedFunction(int x);
}

这里,Foo</*anything except int*/>只有方法,GeneralFunction而Foo<int>只有方法SpecialisedFunction.

按照同样的逻辑,这也是允许的:

template<>
class Foo<float> {
  float GeneralFunction; //here GeneralFunction is a data member, not a method.
}

长话短说,你需要声明你的专业化.