用于继承的 C++ 构造函数

Question

我只是想检查一下，以便从父类继承一个构造函数，这是使用对象组合来实现它的正确方法吗？

我收到了这个错误，它们都指的是我的构造函数。

    C:\Users\user\AppData\Local\Temp\cc0CYtVR.o:SquareImp.cpp:(.text+0x16): undefine
d reference to `vtable for Square'
C:\Users\user\AppData\Local\Temp\cc0CYtVR.o:SquareImp.cpp:(.text+0x79): undefine
d reference to `vtable for Square'
C:\Users\user\AppData\Local\Temp\cc0CYtVR.o:SquareImp.cpp:(.text$_ZN6SquareD1Ev[
Square::~Square()]+0xb): undefined reference to `vtable for Square'
collect2: ld returned 1 exit status

抱歉，这是我的新脚本，因为我已尝试重写它们

ShapeTwoD.h

class ShapeTwoD
{
    protected:
        string name, warpSpace;
        bool containsWarpSpace;
        int xCord, yCord, length, breath;
    public:
        //constructor
        ShapeTwoD();
        ShapeTwoD(string, bool);
};

ShapeTwoD.cpp

ShapeTwoD::ShapeTwoD()
{
    string name = "";
    bool containsWarpSpace = true;
}

ShapeTwoD::ShapeTwoD(string ShapeName, bool warpspace)
{
    name = ShapeName;
    containsWarpSpace = true;
}

方形.h

class Square:public ShapeTwoD
{
    private:
        int xVal,yVal;
        int newlength, newbreath;
        double area;

    public:
        Square();
        Square(string, bool, int, int);
};

正方形.cpp

Square::Square()
{
    xVal = 0;
    yVal = 0;
}

  Square::Square(string ShapeName, bool warpspace, int xval, int yval):ShapeTwoD(ShapeName, warpspace), xVal(xval), yVal(yval)
    {
    xVal = xval;
    YVal = yval;
    cout << "test test" << endl;
}

int main()
{
    Square square;
    cout << "hello world" << endl;
}

Answer 1

是的，还将 xVal 和 yVal 也放入初始化列表中:

Square::Square(string ShapeName, bool square, int xval, int yval):
      ShapeTwoD(ShapeName, square), xVal(xval), yVal(yval)
{
}

同时为 Square() 构造基类:

Square::Square() : ShapeTwoD(..., ...)
{
}