Question

我是C ++编程的新手，我正在尝试执行以下代码，但是它显示错误

没有匹配函数可调用“ Flower :: Flower（）” Rose（字符串n = “没有花”，字符串c =“红色”）：color（c）{}

即使我在Flower类中给了参数构造函数，它仍然说没有匹配的函数调用。

#include <iostream>
#include <string>
using namespace std;


class Flower
{

public:
string name;


Flower (string n):name (n)
  {
  }


void getFlowerName ()
  {
    cout << name << " " << "is" << " ";
} 

};


class Rose:private Flower
{               // Inherit Flower as private
public:
string color;


    /* Initialize name and color data members */ 
Rose (string n = "No flower", string c = "Red"):color (c)
  {
  } 

void getFlowerName (Rose & r)
  {
    cout << r.color << endl;
  } 

    //  using Flower::getFlowerName ;// Allow call to  getFlowerName() method in the base class
};


class Rose:private Flower
{               // Inherit Flower as private
public:
string color;


    /* Initialize name and color data members */ 
Rose (string n = "No flower", string c = "Red"):color (c)
  {
  } 

void getFlowerName (Rose & r)
  {
    cout << r.color << endl;
  } 

using Flower::getFlowerName;    // Allow call to  getFlowerName() method in 
                                   the base class
};

Answer 1

派生类应调用基类：

class Rose : private Flower
{
public:
    std::string color;

    Rose (const std::string& n = "No flower",
          const std::string& c = "Red") : Flower(n), color(c)
    {
    }
// ...
};

目前，您隐式调用默认的Flower的构造函数，但它不存在。

Answer 2

创建Flower::Flower()默认构造函数。

Flower(){ name = NULL; }

使用私有派生类的构造函数初始化基类的数据成员

2 个答案: