继承自专门的自我? [英] inheriting from class of specialized self?
问题描述
这是有效的C ++吗?
template< class category>
,但 gcc-4.5.1 抱怨:
class any_iterator:public any_iterator< void>
{
public:
typedef any_iterator< void> any_iter_void;
any_iterator():any_iter_void(){}
};
模板<>
class any_iterator< void>
{
public:
typedef any_iterator< void> any_iter_void;
any_iterator(){}
void foo(){};
};
int main(){
any_iterator< int>一个;
a.foo(); MSVC10接受它,但在\\上没有错误/警告。
\\ WALL
prog.cpp:3:5:错误:无效使用不完整的类'any_iterator'
prog.cpp:2:11:error :'any_iterator'的声明
prog.cpp:在函数'int main()'中:
prog.cpp:21:11:error:'class any_iterator'has no成员名为'foo'
prog.cpp:在构造函数'any_iterator :: any_iterator()[with category = int]'中:
prog.cpp:20:27:这里
prog.cpp:7:44:error:type'any_iterator'不是'any_iterator'的直接基础
有人可以引用标准显示,如果这应该或不应该编译?我认为这是MSVC中的一个错误。
注意,我知道正确的做法是声明类,专门化根, / em>定义一般情况,这是我将对我的代码,但我想知道哪个编译器是错误的。
解决方案要继承类型,该类型必须是完整的。有点重排解决问题:
模板< class category>
class any_iterator;
模板<>
class any_iterator< void>
{
public:
typedef any_iterator< void> any_iter_void;
any_iterator(){}
void foo(){}
};
template< class category>
class any_iterator:public any_iterator< void>
{
public:
typedef any_iterator< void> any_iter_void;
any_iterator():any_iter_void(){}
};
int main()
{
any_iterator< int&一个;
a.foo();
}
令牌标准报价: / p>
C ++ 11,§10/ 2:
一个 base-type-specifier 应该是一个不是不完全定义类的类类型;这个类被定义的类称为直接基类。
§9.2/ 2: / p>
类在关闭
} $ c时被视为完全定义的对象类型$ c> 。
Is this valid C++?
template<class category> class any_iterator : public any_iterator<void> { public: typedef any_iterator<void> any_iter_void; any_iterator() : any_iter_void() {} }; template<> class any_iterator<void> { public: typedef any_iterator<void> any_iter_void; any_iterator() {} void foo() {}; }; int main() { any_iterator<int> a; a.foo(); }
MSVC10 accepts it with no errors/warnings on
\WALL
, but gcc-4.5.1 complains:prog.cpp:3:5: error: invalid use of incomplete type 'class any_iterator'
prog.cpp:2:11: error: declaration of 'class any_iterator'
prog.cpp: In function 'int main()':
prog.cpp:21:11: error: 'class any_iterator' has no member named 'foo'
prog.cpp: In constructor 'any_iterator::any_iterator() [with category = int]':
prog.cpp:20:27: instantiated from here
prog.cpp:7:44: error: type 'any_iterator' is not a direct base of 'any_iterator'Can someone quote the standard showing if this should or should not compile? I think this is a bug in MSVC.
As a note, I know the correct thing to do is to declare the class, specialize the root, then define the general case, and that's what I'll do to my code, but I was wondering which compiler is wrong here?
解决方案To inherit from a type, that type must be complete. A little rearranging solves things:
template<class category> class any_iterator; template<> class any_iterator<void> { public: typedef any_iterator<void> any_iter_void; any_iterator() { } void foo() { } }; template<class category> class any_iterator : public any_iterator<void> { public: typedef any_iterator<void> any_iter_void; any_iterator() : any_iter_void() { } }; int main() { any_iterator<int> a; a.foo(); }
Token standard quotes:
C++11, §10/2:
The type denoted by a base-type-specifier shall be a class type that is not an incompletely defined class; this class is called a direct base class for the class being defined.
§9.2/2:
A class is considered a completely-defined object type (or complete type) at the closing
}
of the class-specifier.
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