我有一个代码示例,对我来说很奇怪。 使用C ++中的继承,可以使用纯虚函数(也称为Interface)声明指向基类的指针数组,并在其上调用派生成员函数;
class Base {
public:
virtual void call() = 0;
};
class Derived1 : public Base {
public:
void call() override final {
std::wcout << L"derived 1" << std::endl;
}
};
class Derived2 : public Base {
public:
void call() override final {
std::wcout << L"derived 2" << std::endl;
}
};
int main() {
Base* b[2];
b[0] = new Derived1;
b[1] = new Derived2;
for (int i = 0; i < 2; ++i) {
b[i]->call();
}
return 0;
}
给出了:
derived 1
derived 2
按计划进行。 但是当我尝试使用代码示例时,这让我有点困惑:
class Base {
public:
virtual Base* print() = 0;
template<typename T>
Base& operator<<(const T &_val) {
std::wcout << L" d0 << " << _val;
return *this;
}
};
class Derived1 : public Base {
public:
Derived1* print() override final {
return this;
}
template<typename T>
Derived1& operator<<(const T &_val) {
std::wcout << L" d1 << " << _val;
return *this;
}
};
class Derived2 : public Base {
public:
Derived2* print() override final {
return this;
}
template<typename T>
Derived2& operator<<(const T &_val) {
std::wcout << L" d2 << " << _val;
return *this;
}
};
int main() {
Base* b[2];
b[0] = new Derived1;
b[1] = new Derived2;
for (int i = 0; i < 2; ++i) {
std::wcout << typeid(*b[i]->print()).name();
*b[i]->print() << 7 << 7;
std::wcout << std::endl;
}
return 0;
}
输出结果为:
8Derived1 d0 << 7 d0 << 7
8Derived2 d0 << 7 d0 << 7
这意味着只调用 Base的运算符&lt;&lt; (但print()返回类型似乎是正确的。)
问题是为什么它的行为如此?
UPD:
似乎我需要静态多态,这里没有虚函数。但这怎么可能实现呢? 我需要一组不同的Derived类来在operator&lt;&lt;&lt;&lt;在任何数据类型上。 UPD2:
看起来我可以将类型擦除用于运算符&lt;&lt;参数。但是如何在derived的运算符中恢复类型&lt;&lt;然后? (例如,如果我建议使用boost :: any)
答案 0 :(得分:2)
您的operator << ()
不是虚拟的,因此,如果您调用此基类指针的运算符,则始终会调用基类实现。
这是因为它是一种模板方法,模板方法不能virtual
(参见Can a C++ class member function template be virtual?)。
解决方案是为每种受支持的数据类型编写虚拟专用operator << ()
(如果您真的希望它们在继承的类之间有所不同)。
这是更改后的代码:
class Base {
public:
virtual Base* print() = 0;
virtual Base& operator<<( int _val ) = 0;
};
class Derived1 : public Base {
public:
Derived1* print() override final {
return this;
}
Base& operator<<( int _val ) override final {
std::wcout << L" d1 << " << _val;
return *this;
}
};
class Derived2 : public Base {
public:
Derived2* print() override final {
return this;
}
Base& operator<<( int _val ) override final {
std::wcout << L" d2 << " << _val;
return *this;
}
};
int main() {
Base* b[2];
b[0] = new Derived1;
b[1] = new Derived2;
for (int i = 0; i < 2; ++i) {
std::wcout << typeid(*b[i]->print()).name();
*b[i]->print() << 7 << 7;
std::wcout << std::endl;
}
return 0;
}
另一种方法是,仅通过类层次结构(此处为:类名)实现不同的部分,并使用模板运算符进行泛型调用。
#include <iostream>
#include <string>
class Base {
public:
virtual Base* print() = 0;
virtual const wchar_t* className() const = 0;
template<typename T>
Base& operator<<(const T &_val) {
std::wcout << L" " << className() << L" << " << L" << " << _val;
return *this;
}
};
class Derived1 : public Base {
public:
const wchar_t* className() const override final { return L"d1"; }
Derived1* print() override final {
return this;
}
};
class Derived2 : public Base {
public:
const wchar_t* className() const override final { return L"d2"; }
Derived2* print() override final {
return this;
}
};
int main() {
Base* b[2];
b[0] = new Derived1;
b[1] = new Derived2;
for (int i = 0; i < 2; ++i) {
std::wcout << typeid(*b[i]->print()).name();
*b[i]->print() << 7 << 7;
std::wcout << std::endl;
}
return 0;
}
最后一种方法是根据类类型进行RTTI转换,并根据类类型调用专用的operator << ()
。
#include <iostream>
class Derived1;
class Derived2;
class Base {
public:
virtual Base* print() = 0;
};
class Derived1 : public Base {
public:
Derived1* print() override final {
return this;
}
template<typename T>
Derived1& operator<<(const T &_val) {
std::wcout << L" d1 << " << _val;
return *this;
}
};
class Derived2 : public Base {
public:
Derived2* print() override final {
return this;
}
template<typename T>
Derived2& operator<<(const T &_val) {
std::wcout << L" d2 << " << _val;
return *this;
}
};
template<typename T>
Base& operator<<( Base& _base, const T &_val) {
if( typeid( _base ) == typeid( Derived1 ))
return dynamic_cast<Derived1*>(&_base)->operator << (_val);
else
return dynamic_cast<Derived2*>(&_base)->operator << (_val);
}
int main() {
Base* b[2];
b[0] = new Derived1;
b[1] = new Derived2;
for (int i = 0; i < 2; ++i) {
std::wcout << typeid(*b[i]->print()).name();
*b[i]->print() << 7 << 7;
std::wcout << std::endl;
}
return 0;
}