C++ 继承具有多个参数的构造函数 [作业]

Question

我真的希望这是我最近一次问作业问题。在过去的七个小时里，我一直在做这件事，我的所有代码都扔掉了三次，而且我似乎无法克服继承构造函数的问题。我查看了其他问题，遵循了我所看到的每一个建议，但仍然没有骰子。任何和所有的帮助表示赞赏。

头文件:

#ifndef PACKAGE_H
#define PACKAGE_H

struct info
{
    std::string sName;
    std::string sAddr;
    std::string sCity;
    std::string sState;
    std::string sZip;

    std::string rName;
    std::string rAddr;
    std::string rCity;
    std::string rState;
    std::string rZip;
};

class Package
{
 public:
    Package(int c, double w, double co, info *i); //Constructor. Will implement 
                                       //all data types

    ~Package();                            // Default Destructor

    virtual double calculateCost();        // Not pure, can be overloaded

 protected:

    int counter;
    double weight;
    double cost;

    info sendRecieve;                      // Struct for holding tedious info
};

class TwoDayPackage : public Package
{
 public:
    TwoDayPackage(int c, double w, double co, info *i, double a)
    :Package(c, w, co, *i, a);

    double calculateCost();

 private:
    double flatFee;
};

class OvernightPackage : public Package
{
 public:
    OvernightPackage(int c, double w, double co, info *i, double a)
    :Package(c, w, co, *i);

    double calculateCost();

 private:
    double addtlCost;
};

.cpp 定义文件:

#include "Package.h"

using namespace std;

Package::Package(int c, double w, double co, info *i)
{
    cost = c;
    weight = w;
    cost = co;

    sendRecieve.sName = i->sName;
    sendRecieve.sAddr = i->sAddr;
    sendRecieve.sCity = i->sCity;
    sendRecieve.sState = i->sState;
    sendRecieve.sZip = i->sZip;

    sendRecieve.rName = i->rName;
    sendRecieve.rAddr = i->rAddr;
    sendRecieve.rCity = i->rCity;
    sendRecieve.rState = i->rState;
    sendRecieve.rZip = i->rZip;
}

double Package::calculateCost()
{
    return (cost * weight);
}

// end base class definitions

TwoDayPackage::TwoDayPackage(int c, double w, double co, info *i, double a)
    :Package(c, w, co, *i)
{
    flatFee = a;
}

double TwoDayPackage::calculateCost()
{
    return ((cost * weight) + flatFee);
}

//end TwoDayPackage definitions

OvernightPackage::OvernightPackage(int c, double w, double co, info *i, double a)       :Package(c, w, co, *i)
{
    addtlCost = a;
}

double OvernightPackage::calculateCost()
{
    double multi = (addtlCost + cost);

    return (multi * weight);
} 

这是我不断遇到的错误的图片。如果有人可以提供帮助，我将不胜感激。另外，欢迎提出任何建议。预先感谢大家。

Answer 1

Package 的构造函数的最后一个参数是 Info* 而 OvernightPackage 的初始化列表你传递 *i 给它。 i 是 Info* 类型，因此 *i 是 Info& 类型，这不是构造函数所期望的包裹。

对于 TwoDayPackage 的初始化列表中的调用，您将 5 个参数传递给 Package 的构造函数，而它只需要 4 个。