问题描述
请查看以下示例:
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; }
};
这段代码不断给我一个错误,即基类没有默认的构造函数。当我宣布一切都好的时候。但是当我不这样做时,代码不起作用。
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.
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?
How can I declare a copy constructor in derived class without declaring default contructor in base class?
Thnaks。
推荐答案
调用基数的复制构造函数(由编译器生成):
Call the copy-constructor (which is generated by the compiler) of the base:
Derived( const Derived &d ) : Base(d)
{ //^^^^^^^ change this to Derived. Your code is using Base
std::cout << "copy constructor\n";
}
理想情况下,你应该调用编译器生成的基类的复制构造函数。不要想到调用其他构造函数。我认为这是一个坏主意。
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.
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