Question

所有

我在继承链中有3个类（在C ++中），每个类都设置了默认的基础构造函数。但是，链中的第3类抱怨第1个没有匹配的构造函数。下面是构造函数的代码：

class Base
{
    protected:
        int a, b;
    public:
        Base::Base(int first, int second)
        {
            this->a = first;
            this->b = second;
        }
}

class Second : protected Base
{
    protected:
        int c;
    public:
        Second::Second(int first, int second, int third = 2) : Base(first, second)
        {
            this->c = third;
        }
}

class Final : protected Second
{
    private:
        int d;
    public:
        Final::Final(int first, int second, int third = 2) : Second(first, second, third)
        {
        }
}

在编译时我收到错误

＆＃34;在构造函数Final :: Final（int first，int second，int third）

中

没有与Base（）＆＃34;

的匹配调用

为什么这会尝试调用Base（）而不是Base（第一，第二）？

Answer 1

这可能会帮助您构建此类层次结构。

class Base {
protected: 
    int m_a, m_b;

public:
    Base( int first, int second );
    virtual ~Base(){}
};

Base::Base( int first, int second ) :
m_a( first ),
m_b( second ) 
{}

class Second : protected Base {
protected:
    int m_c;
public:
    Second( int first, int second, int third = 2 );
    virtual ~Second(){}
};

Second::Second( int first, int second, int third ) :
Base( first, second ),
m_c( third )
{}

class Final : protected Second {
private:
    int m_d;
public:
   Final( int first, int second, int third = 2 );
   virtual ~Final(){}
};

Final::Final( int first, int second, int third ) :
Second( first, second, third )
{}

如果你想在类中保留你的定义而不是在类声明之外定义它们，那么试试这个

class Base {
protected: 
    int m_a, m_b;

public:
    Base( int first, int second ) : m_a( first ), m_b( second ) {}
    virtual ~Base(){}
};

class Second : protected Base {
protected:
    int m_c;
public:
    Second( int first, int second, int third = 2 ) : 
    Base( first, second ),
    m_c( third ) {}

    virtual ~Second(){}
};

class Final : protected Second {
private:
    int m_d;
public:
   Final( int first, int second, int third = 2 ) :
   Second( first, second, third ) {}

   virtual ~Final(){}
};

优良作法是确保在使用继承时将析构函数声明为虚拟。如果要在类外部定义构造函数或函数，则只需使用类作用域解析运算符，并设置基本成员变量，不需要使用this-＆gt;指针运算符，您可以使用我在定义构造函数时显示的初始化列表。

Answer 2

首先，在类定义结束时缺少select * from pointlayer as p left join linelayer as l on Distance(p.geometry, l.geometry) in ( select MIN(Distance(p.geometry, geometry)) from linelayer where pk in ( select rowid from SpatialIndex where f_table_name = 'linelayer' and search_frame = BuildCircleMbr(X(b.geometry), Y(b.geometry),25)) )。此外，在类中实现成员函数时，不能使用得分运算符;，它在C ++中无效。您可能会发现This有用。

C ++派生类链构造函数错误

2 个答案: