Question

在下面的c ++类中，我已经为构造函数提供了默认参数，以防用户没有提供。但是，当我Point2d1 first(1, 0);中的main()时，我得到一个没有匹配函数调用的错误。我期望将第3个参数默认为0的行为？

·H

#ifndef POINT2D1_H_
#define POINT2D1_H_

class Point2d1 {

private:
    int m_Object_ID;
    double m_x;
    double m_y;


public:
    //Point2d1(int nID);
    Point2d1(int nID, double x, double y);
    virtual ~Point2d1();
    void print() const;
    friend double distanceFrom(const Point2d1& D1, const Point2d1& D2);
};

#endif /* POINT2D1_H_ */

的.cpp

Point2d1::Point2d1(int nID = 0, double x = 0.0, double y = 0.0) : m_Object_ID(nID), m_x(x), m_y(y)
{
    std::cout << "Constructing Point2d object " << nID << '\n';

}

Point2d1::~Point2d1() {
    std::cout << "Destructing Object" << '\n';
}

void Point2d1::print() const
{
    std::cout << "Point2d(" << m_x << ", " << m_y << ")\n";
}

double distanceFrom(const Point2d1& D1, const Point2d1& D2)
{
    double distance = sqrt((D1.m_x - D2.m_x)*(D1.m_x - D2.m_x) + (D1.m_y - D2.m_y)*(D1.m_y - D2.m_y));

    return distance;
}

Answer 1

在头文件的类定义中声明成员函数声明中的默认参数。否则其他编译单元将不知道默认参数。

例如

<强>·H

#ifndef POINT2D1_H_
#define POINT2D1_H_

class Point2d1 {
//...
public:
    //Point2d1(int nID);
    Point2d1(int nID = 0, double x = 0.0, double y = 0.0);
    //...
};

#endif /* POINT2D1_H_ */

和

<强>的.cpp

Point2d1::Point2d1(int nID, double x, double y) : m_Object_ID(nID), m_x(x), m_y(y)
{
    std::cout << "Constructing Point2d object " << nID << '\n';

}

C ++为构造函数提供默认参数

1 个答案: