为什么这不是一个独立的名字？

Question

在类D0中，变量m需要写成this->m，以便使其成为将在基类中查找的依赖名称。

但在类D1中，编译器知道在基类中查找m，而不将m写成this->m。

这怎么可能？为什么D1类中的m不需要写成this->m？

#include <iostream>
#include <string>
template<class T>
class B
{
public:
B(int i) : m(i) {}
T* fb() {std::cout << "B::fb(): m = " << m << "n"; return static_cast<T*>(this); }
protected:
int m;
};
template<class T>
class D0 : public B<T>
{
public:
D0(int i) : B<T>(i) {}
/*
* this-> makes this->m a dependent name so the lookup looks in the base class.  Without this->,
* m would be an independent name and lookup would not check the base class.
*/
T* fd0() {std::cout << "D0::fd0(): m = " << this->m << "n"; return static_cast<T*>(this); }
};
class D1 : public D0<D1>
{
public:
D1(int i) : D0<D1>(i) {}
/*
* D1 doesn't need m qualified by this-> because deriving from D0<D1> somehow
* makes it unnecessary.
*/
D1* fd1() {std::cout << "D1::fd1(): m = " << m << "n"; return this; }
};
int main()
{
std::string s;

D1 d1(2);
d1.fd1()->fd0()->fb()->fd0()->fd1();

std::cout << "Press ENTER to exitn";
std::getline(std::cin, s);
}

Answer 1

不需要详细说明，D1不是一个类模板，它是一个具体的、非模板化的类。因此不需要使用CCD_ 12来访问CCD_。

以下是更多信息。

如果你想复制D1中的错误，下面的例子说明了这一点：

#include <iostream>
#include <string>
template<class T>
class B
{
public:
B(int i) : m(i) {}
T* fb() {std::cout << "B::fb(): m = " << m << "n"; return static_cast<T*>(this); }
protected:
int m;
};
template<class T>
class D0 : public B<T>
{
public:
D0(int i) : B<T>(i) {}
T* fd0() {std::cout << "D0::fd0(): m = " << this->m << "n"; return static_cast<T*>(this); }
};
template <typename T>  // Now let's make this a class template
class D1 : public D0<D1<T>>
{
public:
D1(int i) : D0<D1>(i) {}
D1* fd1() {std::cout << "D1::fd1(): m = " << m << "n"; return this; } // This will now produce an error
};
int main()
{
std::string s;

D1<int> d1(2);
d1.fd1()->fd0()->fb()->fd0()->fd1();

std::cout << "Press ENTER to exitn";
std::getline(std::cin, s);
}

错误：

error: 'm' was not declared in this scope
D1* fd1() {std::cout << "D1::fd1(): m = " << m << "n"; return this; }

现在，一旦D1成为类模板，它就存在错误。