首页 / 知识

关于c ++：GCC问题：使用依赖于模板参数的基类成员

2023-04-12 05:33:00

GCC issue: using a member of a base class that depends on a template argument

以下代码不使用gcc编译，但使用Visual Studio编译：

1
2
3
4
5
6
7
8
9

template <typename T> class A {
public:
T foo;
};

template <typename T> class B: public A < T > {
public:
void bar() { cout << foo << endl; }
};

我收到错误：

test.cpp: In member function ‘void B::bar()’:

test.cpp:11: error: ‘foo’ was not declared in this scope

但它应该是！如果我将bar更改为

1	void bar() { cout << this->foo << endl; }

然后它确实编译，但我不认为我必须这样做。 GCC在这里遵循C ++官方规范中的某些内容，还是仅仅是一个怪癖？

大卫·乔伊纳有历史，这就是原因。

编译B< T >时的问题是它的基类A< T >在编译器中是未知的，是一个模板类，所以编译器无法知道基类中的任何成员。

早期版本通过实际解析基本模板类做了一些推断，但ISO C ++声明这种推断可能导致不应该存在的冲突。

在模板中引用基类成员的解决方案是使用this(与您一样)或专门命名基类：

1
2
3
4
5
6
7
8
9

template <typename T> class A {
public:
T foo;
};

template <typename T> class B: public A < T > {
public:
void bar() { cout << A< T >::foo << endl; }
};

有关gcc手册的更多信息。

哇。 C ++永远不会因为它的古怪而让我感到惊讶。

In a template definition, unqualified names will no longer find members of a dependent base (as specified by [temp.dep]/3 in the C++ standard). For example,

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17

template <typename T> struct B {
int m;
int n;
int f ();
int g ();
};
int n;
int g ();
template <typename T> struct C : B< T > {
void h ()
{
m = 0; // error
f (); // error
n = 0; // ::n is modified
g (); // ::g is called
}
};

You must make the names dependent, e.g. by prefixing them with this->. Here is the corrected definition of C::h,

1
2
3
4
5
6
7

template <typename T> void C< T >::h ()
{
this->m = 0;
this->f ();
this->n = 0
this->g ();
}

As an alternative solution (unfortunately not backwards compatible with GCC 3.3), you may use using declarations instead of this->:

1
2
3
4
5
6
7
8
9
10
11
12
13

template <typename T> struct C : B< T > {
using B< T >::m;
using B< T >::f;
using B< T >::n;
using B< T >::g;
void h ()
{
m = 0;
f ();
n = 0;
g ();
}
};

这只是各种疯狂。谢谢，大卫。

这是他们所指的标准[ISO / IEC 14882：2003]的"temp.dep / 3"部分：

In the definition of a class template or a member of a class template, if a base class of the class template depends on a template-parameter, the base class scope is not examined during unqualified name lookup either at the point of definition of the class template or member or during an instantiation of the class template or member. [Example:

1
2
3
4
5
6
7

typedef double A;
template<class T> class B {
typedef int A;
};
template<class T> struct X : B< T > {
A a; // a has typedouble
};

The type name A in the definition of X< T > binds to the typedef name defined in the global namespace scope, not to the typedef name defined in the base class B< T >. ] [Example:

1
2
3
4
5
6
7
8
9
10
11
12
13

struct A {
struct B { /* ... */ };
int a;
int Y;
};
int a;
template<class T> struct Y : T {
struct B { /* ... */ };
B b; //The B defined in Y
void f(int i) { a = i; } // ::a
Y* p; // Y< T >
};
Y<A> ya;