C/C++教程

C++ Primier Plus(第六版) 第十章 对象和类 编程练习答案

本文主要是介绍C++ Primier Plus(第六版) 第十章 对象和类 编程练习答案,对大家解决编程问题具有一定的参考价值,需要的程序猿们随着小编来一起学习吧!
1. 为复习题5描述的类提供方法定义,并编写一个小程序来演示所有的特性。
本题考查的是类的声明定义以及简单的使用,没有什么难度,样例代码如下:
// account.h -- class defination for Account
#ifndef ACCOUNT_H_
#define ACCOUNT_H_

#include<string>

class BankAccount
{
    private:
        std::string name_;
        std::string acc_num_;
        double deposit_;
    public:
        BankAccount();
        BankAccount(const std::string & name, const std::string & acc_num, double cash);
        ~BankAccount();
        void show_account();
        void deposit_in(double cash);
        void deposit_out(double cash);
};

#endif
// account.cpp -- Account member function
#include <iostream>
#include "account.h"

BankAccount::BankAccount()
{
}

BankAccount::BankAccount(const std::string & name, const std::string & acc_num, double cash)
{
    name_ = name;
    acc_num_ = acc_num;
    deposit_ = cash;
}

BankAccount::~BankAccount()
{
}

void BankAccount::show_account()
{
    using std::cout;
    cout << "Name: " << name_ << ", account number: " 
         << acc_num_ << ", deposit: " << deposit_ << '\n';
}

void BankAccount::deposit_in(double cash)
{
    using std::cout;
    if(cash < 0)
    {
        cout << "The number of cash you deposit is negative; "
             << "transcation is aborted.\n";
    }
    else
        deposit_ += cash;
}

void BankAccount::deposit_out(double cash)
{
    using std::cout;
    if(cash < 0)
    {
        cout << "The number of cash you deposit is negative; "
             << "transcation is aborted.\n";
    }
    else if(cash > deposit_)
    {
        cout << "You can't take the cash morn than you have!"
             << "Transcation is aborted.\n";
    }
    else
        deposit_ -= cash;
}
// accounter.cpp -- test the class account
#include <iostream>
#include "account.h"


int main()
{
    using std::cout;
    using std::cin;
    BankAccount account;
    std::string name;
    std::string acc_num;
    double cash;
    cout << "Enter your name: ";
    getline(cin,name);
    cout << "Enter your account number: ";
    getline(cin,acc_num);
    cout << "Enter the cash: ";
    cin >> cash;
    account = BankAccount(name,acc_num,cash);
    account.show_account();
    cout << "Enter the cash you want to take in: ";
    cin >> cash;
    account.deposit_in(cash);
    account.show_account();
    cout << "Enter the cash you want to take out: ";
    cin >> cash;
    account.deposit_out(cash);
    account.show_account();
    return 0;
}

运行结果如下:
image
2. 下面是一个非常简单的类定义:

class Person
{
private:
    static const LIMIT = 25;
    std::string lname;
    char fname[LIMIT];
public:
    Person(){lname = "";fname[0] = '\0';};
    Person(const std::string & ln, const char * fn = "Heyyou");
// the following methods display lname and fname
    void Show()const;           // firstname lastname format
    void FormalShow() const;    // lastname, firstname format
};

它使用了一个string对象和一个字符数组,让您能够比较他们的用法。请提供未定义的方法的代码,已完成这个类的实现。再编写一个使用这个类的程序,它使用了三种可能的构造函数调用(没有参数、一个参数和两个参数)以及两种显示方法。下面是一个使用这些构造函数和方法的例子。

    Person one;                         // use default constructor
    Person two("Smythecraft");          // use #2 with one default argument
    Person three("Dimwiddy", "Sam");    // use #2, no defaults
    one.Show();
    cout << endl;
    one.FormalShow();
    // etc. for two and three

本题主要考查的是类方法的实现,实现的函数也很简单。比较string与char字符串数组,赋值时一个用=赋值,另一个用strcpy赋值。样例代码如下:

// person.h -- protype class Person
#ifndef PERSON_H_
#define PRESON_H_

#include <string>

class Person
{
private:
    static const int LIMIT = 25;
    std::string lname;
    char fname[LIMIT];
public:
    Person(){lname = "";fname[0] = '\0';};                  // #1
    Person(const std::string & ln, const char * fn = "Heyyou");   // #2
// the following methods display lname and fname
    void Show()const;           // firstname lastname format
    void FormalShow() const;    // lastname, firstname format
};
#endif
// person.cpp -- Person member function
#include <iostream>
#include <cstring>
#include "person.h"

Person::Person(const std::string & ln, const char * fn)
{
    lname = ln;
    strcpy(fname, fn);
}

void Person::Show() const
{
    using std::cout;
    cout << fname << " " << lname;
}

void Person::FormalShow() const
{
    std::cout << lname << ", " << fname << "\n";
}
// ex2_person.cpp -- test the class Person
#include <iostream>
#include "person.h"

int main()
{
    using std::cout;
    using std::endl;
    Person one;                         // use default constructor
    Person two("Smythecraft");          // use #2 with one default argument
    Person three("Dimwiddy", "Sam");    // use #2, no defaults
    one.Show();
    cout << endl;
    one.FormalShow();
    // etc. for two and three
    two.Show();
    cout << endl;
    two.FormalShow();
    three.Show();
    cout << endl;
    three.FormalShow();
}

运行结果如下:
image
3. 完成第9章的编程练习1,但要使用正确的golf类声明替换哪里的代码,用带合适参数的构造函数替换setgolf(golf&,const char *,int),以提供初始值。保留setgolf()的交互版本,但要用构造函数来实现它(例如,setgolf()的代码应该获得数据,将数据传递给构造函数来创建一个临时对象,并将其赋给调用对象,即*this)。
本题的主要内容是将结构修改成类,操作结构的函数改成类的方法。通过修改发现原本函数的参数是结构,改成了对象使用方法。样例代码如下:

// golf.h -- definate Golf class

# ifndef GOLF_H_
# define GOLF_H_

class Golf
{
private:
    static const int Len = 40;
    char fullname[Len];
    int handicap;
public:
    Golf();
    Golf(const char * name, int hc);
    ~Golf();
    int setgolf();
    void set_handicap(int hc);
    void Show() const;
};
# endif
// golf.cpp -- member function of class Golf

#include <iostream>
#include <cstring>
#include "golf.h"

Golf::Golf()
{
}

Golf::Golf(const char * name, int hc)
{
    strcpy(fullname, name);
    handicap = hc;
}
Golf::~Golf()
{
}

int Golf::setgolf()
{
    using std::cout;
    using std::cin;
    using std::endl;
    Golf temp;
    cout << "Enter the users'name: ";
    cin.getline(temp.fullname, Len);
    if(!strcmp(temp.fullname,""))
        return 0;
    cout << "Enter the users'level: ";
    while(!(cin>>temp.handicap))
    {
        cin.clear();
        while(cin.get()!='\n')
            continue;
        cout << "Bad input, please enter a integer: ";
    }
    cin.get();
    *this = temp;
    return 1;
}

void Golf::set_handicap(int hc)
{
    handicap = hc;
}

void Golf::Show() const
{
     using std::cout;
    cout << "Name: " << fullname << ", level: " << handicap << "\n" ;
}
// ex3_usegolf.cpp -- test class Golf
// compile with golf.cpp
#include <iostream>
#include "golf.h"

const int Arsize = 5;
int main()
{
    using std::cout;
    using std::cin;
    using std::endl;
    int hd, number;
    int count = 0;
    Golf gar[Arsize];
    cout << "Enter the Users' name and level(enter empty string to name to quit):\n";
    for(int i = 0; i < Arsize; i++)
    {
        cout << "User #" << i + 1 << ":\n";
        int end_flag = gar[i].setgolf();
        if (end_flag == 0)
            break;
        count++;
    }
    for(int i = 0; i < count; i++)
    {
        cout << "User #" << i + 1 << ":\t";
        gar[i].Show();
    }
    // using handicap

    cout << "Enter the number of user you need to change level(q to quit): ";
    while(cin >> number)
    {
        cout << "The new Level: ";
        cin >> hd;
        gar[number - 1].set_handicap(hd);
        cout << "Enter next number(q to quit): "; 
    }
    cout << "The new golf users list:\n";
    for(int i = 0; i < count; i++)
    {
        cout << "User #" << i + 1 << ":\t";
        gar[i].Show();
    }
    cout << "Bye\n";
    return 0;
}

运行结果如下:
image
4. 完成第9章的编程练习4,但将Sales结构及相关的函数转换成一个类及其方法。用构造函数替换set(sales &, double[], int)函数。用构造函数实现setSales(Sales &)方法的交互版本。将类保留在名称空间SALES中。
本题考查的是在命名空间里定义类和类方法,要改的代码也不是很多。注意在命名空间里定义类方法也需要指名类,确定作用域。样例代码如下:

// sale.h store namespace for sale.cpp
// version class
namespace SALES
{
    const int QUARTERS = 4;
    class Sales
    {
    private:
        double sales[QUARTERS];
        double average;
        double max;
        double min;
    public:
        Sales();
        Sales(const double ar[], int n);
        void ShowSales() const;
    };
}
// sale.cpp -- membet function for Sales in namespace SALES
// compile with ex4_useSales.cpp
#include <iostream>
#include "sale.h"

static double find_min(const double ar[], int n);
static double find_max(const double ar[], int n);
static double getaverage(const double ar[], int n);
static double * find_4less(const double ar[], int n);

namespace SALES
{
    
    Sales::Sales()
    {
        using std::cout;
        using std::cin;
        cout << "Enter four sales:\n";
        for(int i = 0; i < 4; i++)
        {
            cout << "Sales #" << i + 1 << ": ";
            cin >> sales[i];
        }
        min = find_min(sales, 4);
        max = find_max(sales, 4);
        average = getaverage(sales, 4);
    }

    Sales::Sales(const double ar[], int n)
    {
        min = find_min(ar, n);
        max = find_max(ar, n);
        average = getaverage(ar,n);
        if(n <= 4)
        {
            int i;
            for(i = 0; i < n; i++)
                sales[i] = ar[i];
            for(;i < 4; i++)
                sales[i] = 0;
        }
        else
        {
            for(int i = 0; i < 4; i++)
                sales[i] = find_4less(ar, n)[i];
        }
    }

    void Sales::ShowSales() const
    {
        using std::cout;
        cout << "Sales list:\n";
        for(int i = 0; i < 4; i++)
        {
            cout << "Sales #" << i + 1 << ": ";
            cout << sales[i] << "\t";
        }
        cout << "\n";
        cout << "Average = " << average << ", max = " << max;
        cout << ", min = " << min << "\n";
    }
}

double find_min(const double ar[], int n)
{
    double min = ar[0];
    for(int i = 0; i < n; i++)
        min = min < ar[i] ? min : ar[i];
    return min;
}

double find_max(const double ar[], int n)
{
    double max = ar[0];
    for(int i = 0; i < n; i++)
        max = max < ar[i] ? ar[i] : max;
    return max;
}

double getaverage(const double ar[], int n)
{
    double sum = 0;
    for(int i = 0; i < n; i++)
        sum += ar[i];
    return sum / n;
}
static double * find_4less(const double ar[], int n)
{
    double * lesser4 = new double[4];
    double lesser;
    lesser4[0] = find_min(ar, n);
    for(int i = 1; i < 4; i++)
    {
        lesser = find_max(ar, n);
        for(int j = 0; j < n; j++)
        {
            if(ar[j] > lesser4[i-1])
                lesser = lesser < ar[j] ? lesser : ar[j]; 
        }
        lesser4[i] = lesser;
    }
    return lesser4;
}
// ex4_useSales.cpp -- using namespace and class
// compile with sale.cpp

#include <iostream>
#include "sale.h"

int main()
{
    using namespace SALES;
    using std::cout;
    Sales s1;
    double arr[6] = {1111.1, 666.6, 999.9, 333.3, 2222.2, 1234.5};
    Sales s2 = Sales(arr, 6);
    s1.ShowSales();
    s2.ShowSales();
    cout << "Bye\n";
    return 0;
}

运行结果如下:
image
5. 考虑下面的结构声明:

struct customer{
	char fullname[35];
	double payment;
}

编写一个程序,它从栈中添加和删除custome结构(栈用Stack类声明表示)。每次customer结构被删除时,其payment的值都将被加入到总数中,并报告总数。注意:应该可以直接使用Stack类而不作修改;只需修改typedef声明,使Item的类型为customer,而不是unsigned long即可。

本题在编写过程中遇到了不少问题,首先是不理解stack的pop与push程序,栈顶元素是空着的,压入栈时将元素赋给栈顶元素,然后指针top++。弹出时,首先--top,然后将top指向元素的值弹出。
使用数组实现时:
push stack[top++] = value;该程序等价于stack[top] = value;top++;
pop value = stack[--top];该程序等价于--top;value = stack[top];
样例代码如下:

// stack.h -- Class Stack protype and  member function protype
#ifndef STACK_H_
#define STACK_H_

const int NSIZE = 35;

struct customer
{
    char fullname[NSIZE];
    double payment;
};

typedef struct customer Item;

class Stack
{
private:
    static const int MAX = 10; // also enum{MAX = 10};
    Item items[MAX];
    int top;
public:
    Stack();
    bool isempty() const;
    bool isfull() const;
    bool push(const Item & item);
    bool pop(Item & item);
};
#endif
// stack.cpp -- member function for stack class
// compile with ex4.cpp

#include <cstring>
#include "stack.h"

Stack::Stack()
{
    top = 0;
}

bool Stack::isempty() const
{
    return top == 0;
}

bool Stack::isfull() const
{
    return top == MAX;
}

bool Stack::push(const Item & item)
{
    if(isfull())
        return false;
    strcpy(items[top].fullname,item.fullname);
    items[top++].payment = item.payment;
    return true;
}

bool Stack::pop(Item & item)
{
    if(top > 0)
    {
        item = items[--top];
        return true;
    }
    else
        return false;
}
// ex4_usestack.cpp -- use stack class to store the custormer
// compile with stack.cpp
#include <iostream>
#include <cstring>
#include "stack.h"

int main()
{
    using std::cin;
    using std::cout;
    using std::endl;
    Stack st;
    customer ct;
    double sum_pm = 0;
    int count = 0;
    cout << "Enter the customer name(enter empty string to quit): ";
    cin.getline(ct.fullname,NSIZE);
    while (strcmp(ct.fullname, ""))
    {
        cout << "Enter the payment: ";
        while(!(cin >> ct.payment))
        {
            cin.clear();
            while (cin.get() != '\n')
                continue;
            cout << "Bad input, please input a number: ";
        }
        cin.get();
        st.push(ct);
        if(st.isfull())
            break;
        cout << "Enter next customer name(enter empty string to quit): ";
        cin.getline(ct.fullname,NSIZE);
    }
    while(!st.isempty())
    {
        ++count;
        st.pop(ct);
        sum_pm += ct.payment;
        cout << count  << " customers' total payment is " << sum_pm << endl;
    }
    cout << "Bye\n";
    return 0;
}

运行结果如下:
image
6. 下面是一个类声明:

class Move
{
private:
    double x;
    double y;
public:
    Move(double a = 0.0,double b = 0.0);    // sets x, y to a, b
    void showmove() const;                  // shows current x, y values
    Move add(const Move & m) const;
// this function adds x of m to x of invoking object to get new x,
// adds y to y of invokeing object to get new y, create a new
// move object initialized to new x, y values and returns it
    void reset(double a = 0.0, double b = 0.0);  // resets x,t to a,b
};

请提供成员函数的定义和测试这个类的程序
本题比较简单,就是声明类和测试类,样例代码如下:

// move.h -- defination of Move class for move.cpp
// version

#ifndef MOVE_H_
#define MOVE_H_

class Move
{
private:
    double x;
    double y;
public:
    Move(double a = 0.0,double b = 0.0);    // sets x, y to a, b
    void showmove() const;                  // shows current x, y values
    Move add(const Move & m) const;
// this function adds x of m to x of invoking object to get new x,
// adds y to y of invokeing object to get new y, create a new
// move object initialized to new x, y values and returns it
    void reset(double a = 0.0, double b = 0.0);  // resets x,t to a,b
};
#endif
// move.cpp -- member function of class Move
// compile with ex6_usemove.cpp

#include <iostream>
#include "move.h"

Move::Move(double a, double b)
{
    x = a;
    y = b;
}

void Move::showmove() const
{
    std::cout << "x = " << x << ", y = " << y << '\n';
}

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

void Move::reset(double a , double b)
{
    x = a;
    y = b;
}
// ex6_usemove.cpp -- test class Move
// compile with move.cpp

#include <iostream>
#include "move.h"

int main()
{
    using std::cin;
    using std::cout;
    Move point;
    double a, b;
    Move mv;
    Move result;
    cout << "Enter the point of x,y position: ";
    cin >> a >> b;
    point.reset(a,b);
    point.showmove();
    cout << "Enter the move x y: ";
    cin >> a >> b;
    mv.reset(a,b);
    result = point.add(mv);
    cout << "After move, the point of position is ";
    result.showmove();
    return 0;
}

运行结果如下:
image
7. Betelgeusean plorg 有这些特征。
数据:

  • plorg的名称不超过19个字符;
  • plorg有满意指数(CI),这是一个整数。

操作:

  • 新的plorg将有名称,其CI值为50;
  • plorg的CI可以修改;
  • plorg可以报告其名称和CI;
  • plorg的默认名称为"Plorga"。

请编写一个Plorg类声明(包括成员函数和成员函数原型)来表示plorg,并编写成员函数的函数定义。然后编写一个小程序,以演示Plorg类的所有特性。
本题比较简单,首先将题目中的文字转换成代码实现,这些没有问题,在写测试主函数的时候,遇到了问题,由于cin >> CI 的优先级比!低,因此又该有括号!(cin >> CI),样例代码如下:

// plorg.h -- defination of class Plorg
// version 00

#ifndef PLORG_H_
#define PLORG_H_
const int NSIZE = 20;
class Plorg
{
private:
    char name_[NSIZE];
    int CI_;
public:
    Plorg();
    Plorg(const char * name, int CI = 50);
    void setCI(int CI);
    void ShowPlorg() const;
    void setname(const char * name = "Plorga");
};
#endif
// plorg.cpp -- member funciton of class Plorg
// compile with ex7_useplorg.cpp

#include <iostream>
#include <cstring>
#include "plorg.h"

Plorg::Plorg()
{
}

Plorg::Plorg(const char * name, int CI)
{
    strcpy(name_, name);
    CI_ = CI;
}

void Plorg::setCI(int CI)
{
    CI_ = CI;
}

void Plorg::ShowPlorg() const
{
    std::cout << "Plorg name: " << name_ << ", CI = " << CI_ << '\n';
}

void Plorg::setname(const char * name)
{
    strcpy(name_,name);
}
// ex7_useplorg.cpp -- test class Plorg
// compile with plorg.cpp

#include <iostream>
#include <cstring>
#include "plorg.h"

const int PSize = 10;
int main()
{
    using std::cout;
    using std::cin;
    Plorg plorg[PSize];
    Plorg pl = Plorg("Jobs Sam",60);
    pl.ShowPlorg();
    char name[NSIZE];
    int CI;
    int count = 0;
    for(int i = 0; i < PSize; i++)
    {
        cout << "The plorg #" << i + 1 << ":\n";
        cout << "Enter the name(enter empty string to quit): ";
        cin.getline(name, NSIZE);
        if(!strcmp(name,""))
            break;
        count++;
        cout << "Enter the CI: ";
        while(!(cin >> CI)) // () is must
        {
            cin.clear();
            while(cin.get() != '\n')
                continue;
            cout << "Bad input, please input a integer: ";
        }
        cin.get();
        plorg[i].setname(name);
        plorg[i].setCI(CI);
    }
    cout << "Plorg List:\n" ;
    for(int i = 0; i < count; i++)
        plorg[i].ShowPlorg();

    cout << "Bye\n";

}

运行结果如下:
image
8. 可以将简单列表描述成下面这样:

  • 可以存储0或多个某种类型的列表;
  • 可创建空列表;
  • 可在列表中添加数据项;
  • 可确定列表是否为空;
  • 可确定列表是否为满;
  • 可以访问列表中每一个数据项,并对他执行某种操作。

可以看到,这个列表确实简单,例如它不允许插入或删除数据项。您应该提供头文件list.h和实现文件list.cpp,前者包含类定义,后者包含类方法的实现。您还应该创建一个简单的程序来使用这个类。
该列表的规范很简单,这主要旨在简化这个编程练习。可以选择使用数组或链表来实现该列表,但公有接口不应依赖于所做的选择,也就是说,公有接口不应有数组索引、节点指针等。应使用通用概念来表达创建列表、在列表中添加数据项等操作。对于访问数据项以及执行操作,通常应使用将函数指针作为参数的函数来处理:
void visit(void (*pf) (Item &))
其中,pf指向一个将Item引用作为参数的函数(不是成员函数),Item是列表中数据项的类型。visit()函数将该函数用于列表中的每个数据项。
本题由于笔者目前没有学过链表,因此没有使用链表,使用了数组来存储,编写程序时存在问题,visit函数,有问题,后面理解了函数作为参数之后,将程序写出来了,测试程序比较简陋,以后学了链表再来改进这个程序,也体会到了面向对象编程的好处。代码如下:

// list.h -- protype class list
// version 00

#ifndef LIST_H_
#define LIST_H_
typedef double Item;
class List
{
private:
    enum{MAX = 10};
    Item items[MAX];
    int header;
public:
    List();
    List(const Item * aritem, int n);
    bool add_data(const Item & item);
    bool isempty() const;
    bool isfull() const;
    void visit(void (*pf) (Item & item));
};
#endif
// list.cpp -- member function of class List
// compile with ex8_uselist.cpp

#include <iostream>
#include "list.h"

List::List()
{
    header = 0;
}

List::List(const Item * aritem, int n)
{
    for(header = 0; header < n; header++)
    {
        items[header] = aritem[header]; 
    }
}

bool List::add_data(const Item & item)
{
    if(header < MAX)
    {
        items[header++] = item;
        return true;
    }
    else
        return false;
}

bool List::isempty() const
{
    return header == 0;
}

bool List::isfull() const
{
    return header == MAX;
}

void List::visit(void (*pf) (Item & item))
{
    for(int i = 0; i < header; i++)
        (*pf)(items[i]);
}
// ex8_uselist.cpp -- test class List
// compile with list.cpp

#include <iostream>
#include "list.h"
void Show(Item & item);
int main()
{
    using std::cout;
    Item num;
    List ld1;
    double arr[10] = {1.1, 1.2, 1.3, 1.4, 1.5, 1.6, 1.7, 1.8, 1.9, 2.0};
    if(ld1.isempty())
        cout << "List of double is empty\n";
    List ld2 = List(arr,10);
    if(ld2.isfull())
        cout << "List of double is full\n";
    ld2.visit(Show);
    cout << '\n';
    cout << "Before add data:\n";
    ld1.visit(Show);
    cout << '\n';
    cout << "Enter the number you want to add: ";
    std::cin >> num;
    ld1.add_data(num);
    cout << "After add data:\n";
    ld1.visit(Show);
    cout << '\n';
    cout << "Bye\n";
    return 0;
}
void Show(Item & item)
{
    std::cout << item << " "; 
}

运行结果如下:
image

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