C++ 如何在派生类中声明复制构造函数，而基类中没有默认构造函数？
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How to declare copy constructor in derived class, without default construcor in base?
提问by unresolved_external
Please take a look on the following example:
请看下面的例子：
class Base
{
protected:
int m_nValue;
public:
Base(int nValue)
: m_nValue(nValue)
{
}
const char* GetName() { return "Base"; }
int GetValue() { return m_nValue; }
};
class Derived: public Base
{
public:
Derived(int nValue)
: Base(nValue)
{
}
Derived( const Base &d ){
std::cout << "copy constructor\n";
}
const char* GetName() { return "Derived"; }
int GetValueDoubled() { return m_nValue * 2; }
};
This code keeps throwing me an error that there are no default contructor for base class. When I declare it everything is ok. But when i dont, code does not work.
这段代码不断向我抛出一个错误，即基类没有默认构造函数。当我宣布它时一切正常。但是当我不这样做时，代码不起作用。
How can I declare a copy constructor in derived class without declaring default contructor in base class?
如何在派生类中声明复制构造函数而不在基类中声明默认构造函数？
Thnaks.
纳克斯。
回答by Nawaz
Call the copy-constructor (which is generated by the compiler) of the base:
调用基类的复制构造函数（由编译器生成）：
Derived( const Derived &d ) : Base(d)
{ //^^^^^^^ change this to Derived. Your code is using Base
std::cout << "copy constructor\n";
}
And ideally, you should call the compiler generated copy-constructor of the base. Don't think of calling the other constructor. I think that would be a bad idea.
理想情况下，您应该调用编译器生成的基类复制构造函数。不要想着调用另一个构造函数。我认为那将是一个坏主意。
回答by PlasmaHH
You can (and should) call the copy ctor of the base class, like:
您可以（并且应该）调用基类的复制构造函数，例如：
Derived( const Derived &d ) :
Base(d)
{
std::cout << "copy constructor\n";
}
Note that I turned the Base parameter into a Derived parameter, since only that is called a copy ctor. But maybe you didn't really wanted a copy ctor...
请注意，我将 Base 参数转换为 Derived 参数，因为只有它才称为复制构造函数。但也许你并不真正想要一个复制 ctor...
