[英]trouble with operator overloading in baseclass
Hi I'm having some trouble with inhertance and operator overloading and I'm hoping you guys can give me some clarity. 嗨,我在继承和运算符重载方面遇到了麻烦,希望你们能给我一些清晰度。
I have the following classes: 我有以下课程:
template<typename Type>
class Predicate{
public:
Predicate() {};
virtual ~Predicate(){};
virtual bool operator()(const Type & value) = 0;
virtual bool operator()(const Type * value){ //<-- this is the operator thats not working
return (*this)(*value);
};
};
template<typename Type>
class Always : public Predicate<Type>{
public:
bool operator()(const Type & value){return true;}
~Always(){};
};
Now I want all my predicates to accept both references and pointers, but when I test the classes in: 现在,我希望所有谓词都接受引用和指针,但是当我在以下类中测试类时:
int main(){
Always<int> a;
int i = 1000;
a(&i);
system("pause");
return 1;
}
I receive the following error: 我收到以下错误:
test.cpp: In function 'int main()':
test.cpp:10:6: error: invalid conversion from 'int*' to 'int' [-fpermissive]
a(&i);
^
In file included from test.cpp:2:0:
predicates.h:22:7: error: initializing argument 1 of 'bool Always<Type>::operator()(const Type&) [with Type = int]' [-fpermissive]
bool operator()(const Type & value){return true;}
This is because when you are declaring: 这是因为在声明时:
bool operator()(const Type & value){return true;}
in the subclass, you are hiding / shadowing any other overload of the operator in the superclass. 在子类中,您要隐藏 / 隐藏超类中运算符的任何其他重载。
If you add: 如果添加:
using Predicate<Type>::operator();
in the subclass, everything will work fine. 在子类中,一切正常。
On a side note, I think that allowing both const&
and const*
is a design smell. 从侧面讲,我认为同时允许
const&
和const*
是一种设计味道。 You should just allow the const&
version and let the user of your class do *ptr
if they have a ptr
pointer. 您应该只允许
const&
版本,并允许类的用户执行*ptr
如果他们具有ptr
指针)。
Templates and operator overloading obfuscate the real problem here. 模板和运算符重载掩盖了这里的实际问题。 Look at this small piece of code which yields the same error:
看一下产生相同错误的这段小代码:
void f(int &);
int main()
{
int *ptr;
f(ptr);
}
The compiler won't let you pass a pointer where a reference is expected. 编译器不会让你通过凡在预期的指针。 This is what you try to do with your derived class.
这就是您尝试对派生类进行的操作。 As you operate on a concrete
Always
, the base versions of operator()
are not considered. 当你对一个具体的操作
Always
,的基本版本operator()
不考虑。
Look how the situation changes when you operate instead on a pointer (or reference) to the base class: 查看当您对基类的指针(或引用)进行操作时,情况如何变化:
int main(){
Predicate<int> *ptr = new Always<int>;
int i = 1000;
(*ptr)(&i);
delete ptr;
}
This compiles fine because the base-class operators are now considered for overload resolution. 编译良好,因为现在考虑将基类运算符用于重载解析。 But this is just to make you understand the problem better.
但这只是为了使您更好地理解问题。 The solution is to apply the Non-Virtual Interface Idiom .
该解决方案是应用的非虚拟接口成语 。 Make your operators non-virtual and implement them in terms of private virtual functions:
使您的操作员不是虚拟的,并根据私有虚拟功能实施它们:
template<typename Type>
class Predicate{
public:
Predicate() {};
virtual ~Predicate(){};
bool operator()(const Type & value) { return operatorImpl(value); }
bool operator()(const Type * value) { return operatorImpl(value); }
private:
virtual bool operatorImpl(const Type & value) = 0;
virtual bool operatorImpl(const Type * value) {
return (*this)(*value);
}
};
template<typename Type>
class Always : public Predicate<Type>{
public:
~Always(){};
private:
bool operatorImpl(const Type & value){return true;}
};
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