问题描述
为什么C ++编译器不能识别 g()
和 b
$ c>如下代码中所示的超类:
Why can't a C++ compiler recognize that g()
and b
are inherited members of Superclass
as seen in this code:
template<typename T> struct Superclass {
protected:
int b;
void g() {}
};
template<typename T> struct Subclass : public Superclass<T> {
void f() {
g(); // compiler error: uncategorized
b = 3; // compiler error: unrecognized
}
};
如果我简化子类
子类< int>
然后它编译。当完全限定 g()
为超类 :: g()
和超类< T> :: b
。我使用LLVM GCC 4.2。
If I simplify Subclass
and just inherit from Subclass<int>
then it compiles. It also compiles when fully qualifying g()
as Superclass<T>::g()
and Superclass<T>::b
. I'm using LLVM GCC 4.2.
注意:如果我使 g()
和
Note: If I make g()
and b
public in the superclass it still fails with same error.
推荐答案
这可以通过修改使用使用
将名称拉入当前范围:
This can be amended by pulling the names into the current scope using using
:
template<typename T> struct Subclass : public Superclass<T> {
using Superclass<T>::b;
using Superclass<T>::g;
void f() {
g();
b = 3;
}
};
或通过 this
指针访问:
template<typename T> struct Subclass : public Superclass<T> {
void f() {
this->g();
this->b = 3;
}
};
或者,您已经注意到,通过限定全名。
Or, as you’ve already noticed, by qualifying the full name.
这是必要的原因是C ++不考虑名称解析的超类模板(因为它们是依赖名称,不考虑依赖名称)。当您使用超类时,因为它不是模板(它是模板的一个实例化),因此它的嵌套名称不是依赖名称。
The reason why this is necessary is that C++ doesn’t consider superclass templates for name resolution (because then they are dependent names and dependent names are not considered). It works when you use Superclass<int>
because that’s not a template (it’s an instantiation of a template) and thus its nested names are not dependent names.
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