無法訪問在課堂上聲明的私人成員

Question

我有一段代碼，它是方法定義

Move add(const Move & m) {
    Move temp;
    temp.x+= (m.x +this-x);
    temp.y+= (m.y + this->y);
    return temp;
}

這是類聲明

class Move
{
private:
    double x;
    double y;
public:
    Move(double a=0,double b=0);
    void showMove() const;
    Move add(const Move & m) const;
    void reset(double a=0,double b=0);
};

它說

1>c:\users\filip\dysk google\c++\consoleapplication9\move.cpp(18): error C2248: 'Move::x' : cannot access private member declared in class 'Move'
1>          c:\users\filip\dysk google\c++\consoleapplication9\move.h(7) : see declaration of 'Move::x'
1>          c:\users\filip\dysk google\c++\consoleapplication9\move.h(5) : see declaration of 'Move'
1>          c:\users\filip\dysk google\c++\consoleapplication9\move.h(7) : see declaration of 'Move::x'
1>          c:\users\filip\dysk google\c++\consoleapplication9\move.h(5) : see declaration of 'Move'
1>c:\users\filip\dysk google\c++\consoleapplication9\move.cpp(18): error C2355: 'this' : can only be referenced inside non-static member functions
1>c:\users\filip\dysk google\c++\consoleapplication9\move.cpp(18): error C2227: left of '->x' must point to class/struct/union/generic type

與Move :: y相同。 Any1知道那是什么嗎？

Answer 1

您需要在Move類范圍內定義add ：

Move Move::add(const Move & m) const {
    Move temp;
    temp.x+= (m.x +this-x);
    temp.y+= (m.y + this->y);
    return temp;
}

否則，它將被解釋為非成員函數，無法訪問Move的非公共成員。

請注意，假設兩個參數構造函數設置了x和y ，則可以簡化代碼：

Move Move::add(const Move & m) const {
    return Move(m.x + this-x, m.y + this->y);
}