[英]Finding names in templatized base classes in C++
I'm reading Effective C++ 3rd Edition, item 43 "Know how to access names in templatized base classes". 我正在阅读有效的C ++ 3rd Edition,项目43“知道如何访问模板化基类中的名称”。
template<typename T>
class B {
T i;
};
template<typename T>
class D: public B<T> {
public:
void Foo() {
T a = B<T>::i;
}
};
int main() {
D<int> d;
}
For the above codes, I know if the B<T>::
is not added before i
in D::Foo()
, compilers will complain " i
was not declared in this scope". 对于以上代码,我知道是否在
D::Foo()
i
之前未添加B<T>::
,编译器会抱怨“未在此范围内声明i
”。 (But it didn't complain i
is private in B
.) (但是它并没有抱怨
i
在B
是私人的。)
However, if T i;
但是,如果
T i;
is not declared in B, like the following, the compiling goes well. 没有在B中声明,如下所示,编译进行顺利。
template<typename T>
class B {
};
template<typename T>
class D: public B<T> {
public:
void Foo() {
T a = B<T>::i;
}
};
int main() {
D<int> d;
}
Compilers are defaulted not to find names in templatized base classes. 默认情况下,编译器不在模板化的基类中找到名称。 But why they still don't do even I told them?
但是,为什么我甚至告诉他们,他们仍然不这样做?
But why they still don't do even I told them?
但是,为什么我甚至告诉他们,他们仍然不这样做?
Because the member function Foo
is not used, then it's not instantiated at all. 因为没有使用成员函数
Foo
,所以根本不会实例化它。
This applies to the members of the class template: unless the member is used in the program, it is not instantiated, and does not require a definition.
这适用于类模板的成员:除非在程序中使用了该成员,否则不会实例化该成员,并且不需要定义。
You might get an error if Foo
is called, like 如果调用
Foo
可能会出错 ,例如
D<int> d;
d.Foo();
BTW BTW
But it didn't complain
i
is private inB
.但这并没有抱怨
i
在B
是私人的。
Because accessibility check is performed after name lookup. 因为在名称查找之后执行了可访问性检查。 The name
i
is not found, then accessibility of nothing could be checked. 没有找到
i
的名字,则无法检查任何内容的可访问性。
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