为什么编译器选择模板参数列表中的基类构造函数? [英] Why does the compiler select the base class constructor inside the template argument list?
问题描述
后续问题要这一个。
基本上,在下面的代码中,编译器为什么认为 B
在 A
$ c> C 的构造函数是指 B
基类的(不可访问)构造函数
$ b b
struct B {};
template< typename T>
struct A:private T {};
struct C:public A< B> {
C(A< B>); // ERROR HERE
};
Ideone上的实例。 输出:
prog.cpp:1:9:错误:'struct BB :: B'无法访问
prog.cpp:7:7:错误:在此上下文中
请注意, A< B *>
, A 或甚至
A< const B>
。还要注意,MSVC10,GCC 4.7和Clang 3.1 ToT中的三个会出错,所以它必须是C ++规范中的一些东西。 是什么?
标准允许注入的类名称比原始名称。这甚至在§11.1/ 5的注释中提到,连同一个例子:
[在派生类中,基类名称的查找将
在
中找到注入类名,而不是基类的名称,在其中声明它。 inject-class-name可能比在声明
的作用域中的基类的名称
更不可访问。
< p $ p>class A {};
class B:private A {};
class C:public B {
A * p; //错误:inject-class-name A是不可访问的
:: A * q; // OK
};
- end example ]
< blockquote>
访问
A
不合格使用注入的名称,这是不可访问的,因为它来自私人继承。访问A
限定使用已声明的名称,可在全局范围中访问。Follow-up question to this one.
Basically, in the following code, why does the compiler think that the
B
insideA<B>
inC
s constructor refer to the (inaccessible) constructor of theB
base class?struct B{}; template <typename T> struct A : private T{}; struct C : public A<B>{ C(A<B>); // ERROR HERE };
Live example on Ideone. Output:
prog.cpp:1:9: error: 'struct B B::B' is inaccessible
prog.cpp:7:7: error: within this contextNote that the same error pops up if you change the constructor argument to
A<B*>
,A<B&>
or evenA<const B>
. Also note that three of MSVC10, GCC 4.7 and Clang 3.1 ToT will error out, so it must be something in the C++ spec. What is it?解决方案The standard allows injected class names to be less accessible than the original names. This is even mentioned in a note in §11.1/5, together with an example:
[ Note: In a derived class, the lookup of a base class name will find the injected-class-name instead of the name of the base class in the scope in which it was declared. The injected-class-name might be less accessible than the name of the base class in the scope in which it was declared. —end note ]
[ Example:
class A { }; class B : private A { }; class C : public B { A *p; // error: injected-class-name A is inaccessible ::A *q; // OK };
—end example ]
Accessing
A
unqualified uses the injected name, which is not accessible because it comes from private inheritance. AccessingA
qualified uses the declared name, which is accessible in the global scope.这篇关于为什么编译器选择模板参数列表中的基类构造函数?的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!