冒泡排序不适用于对 C++ 中的动态对象数组进行排序

Posted 2023-02-21

【中文标题】冒泡排序不适用于对 C++ 中的动态对象数组进行排序

【英文标题】：Bubble sorting doesn't work to sort dynamic array of objects in C++

【发布时间】：2017-12-08 20:04:14

【问题描述】：

我的 City 类具有以下私有数据：城市名称（char 类型）、宽度（double）、长度（double）和高度（double）。我必须制作动态数组，默认情况下由构造函数插入 - City()，当程序启动时。然后程序使用方法 output() 并打印插入的城市数组。mA 是我的城市对象。我应该使用气泡排序以按城市的长度对城市进行排序。我有复制构造函数，运算符 = 并且我必须将 double max（这是变量，在冒泡排序中用于存储当前最大值）转换为 City 类型，为此我使用了带有一个的构造函数参数：City(double max)。问题是排序不起作用。我认为问题出在我对带有一个参数的构造函数的定义中（将 double 类型转换为 city 类型）。

#include "stdafx.h"
#include<iostream>
#include<math.h>
#include <algorithm> 
using namespace std;

class City
private: char *name;
         double width;
         double length;
         double height;
public:

    void Output();
    City();
    ~City();
    City(double max)
        name = "";
        width = 0;
        length = max;
        height = 0;
    
    double GetLength()
    
        return length;
    
    double GetWidth() return width; 
    double GetHeight() return height; 
    char GetName() return *name; 
    City(const City& that)
    
        name = new char[strlen(that.name) + 1];
        for (int i = 0; i <= strlen(that.name); i++)
            name[i] = that.name[i];
        //strcpy(name, that.name);
        width = that.width;
        length = that.length;
        height = that.height;
    


    City& operator=(const City that)
    
        name = that.name;
        width = that.width;
        length = that.length;
        height = that.height;
        return*this;
    
;
City::City()

    char ime[20];
    cout << "Name= ";
    cin >> ime;
    name = new char[strlen(ime) + 1];
    for (int i = 0; i <= strlen(ime); i++)
        name[i] = ime[i];

    cout << "Width= ";
    cin >> width;
    cout << "Length= ";
    cin >> length;
    cout << "Height= ";
    cin >> height;


void City::Output()

    cout << "Name is: " << name << endl;
    cout << " Width is: " << width << " deg" << endl;;
    cout << " Length is: " << length << " deg" << endl;
    cout << " Height is: " << height << " m" << endl;
    return;

City::~City()

    cout << " " << endl;
    cout << "Destructor of City!" << endl;
    delete[] name;



int main()


    int n;
    City *mA;
    cout << "Input number of cities: " << endl;
    cin >> n;
    mA = new City[n];
    for (int j = 0; j < n; j++)
    
        mA[j].Output();
    
    cout << "Cities from west to east, sorted by their length" << endl;


    double max = mA[0].GetLength();
    for (int j = 1; j<n; j++)
    
        if (mA[j - 1].GetLength()>mA[j].GetLength())
        
            max = mA[j - 1].GetLength();

            mA[j - 1] = mA[j];
            mA[j] = max;
        
    
    for (int j = 0; j < n; j++)
    
        mA[j].Output();
    

    delete[]mA;
    return 0;

【问题讨论】：

“Doesn't work”不是错误描述，您希望它做什么以及它会做什么？

冒泡排序有两个嵌套循环。

您应该使用std::swap 交换值。现在您的mA[j] = max; 每次都会从max 创建一个新对象（使用您怀疑的构造函数）。如果您不想要隐式转换，可以使用 explicit City(double max)。

name = ""; 为常量数据分配一个非常量指针。对此要非常小心。例如，想想当这个对象被销毁并运行delete[] name; 时会发生什么。考虑改为设置name = nullptr;。如果您尝试滥用 name，您将收到更多有用的错误消息。

使用 std::string，而不是 char 数组。

【参考方案1】：

City::City()

    char ime[20];
    cout << "Name= ";
    cin >> ime;
    name = new char[strlen(ime) + 1];
    for (int i = 0; i <= strlen(ime); i++)
        name[i] = ime[i];
    ...

首先你必须修复构造函数。虽然不一定是错的，但你不应该在构造函数中要求输入。改为为Input() 添加一个单独的函数

City& operator=(const City that)

    name = that.name;
    ...
    return*this;

这个赋值运算符是错误的。 name 是一个指针，您不想分配指针（不在这种情况下）。您应该使用与以前相同的方法来复制名称：

name = new char[strlen(ime) + 1];
strcpy(name, ime);

冒泡排序应该像下面的方法一样完成。另外，您添加了#included <string>。你应该改用std::string！

#include <iostream>
#include <string>

using namespace std;

class City

private:
    char *name;
    double width;
    double length;
    double height;
public:
    City()
    
        name = nullptr;
        width = 0;
        length = 0;
        height = 0;
    

    City(const City& that)
    
        name = new char[strlen(that.name) + 1];
        strcpy(name, that.name);
        width = that.width;
        length = that.length;
        height = that.height;
    

    City& operator=(const City that)
    
        name = new char[strlen(that.name) + 1];
        strcpy(name, that.name);
        width = that.width;
        length = that.length;
        height = that.height;
        return*this;
    

    ~City()
    
        delete[] name;
    

    void Input()
    
        char buffer[100];
        cout << "Name= ";
        cin >> buffer;
        name = new char[strlen(buffer) + 1];
        strcpy(name, buffer);
        cout << "Width= ";
        cin >> width;
        cout << "Length= ";
        cin >> length;
        cout << "Height= ";
        cin >> height;
    

    void Output()
    
        cout << "Name is: " << name << ", " << length << endl;
        cout << " Width is: " << width << " deg" << endl;
        cout << " Length is: " << length << " deg" << endl;
        cout << " Height is: " << height << " m" << endl << endl;
    

    double GetLength()  return length; 
    double GetWidth()  return width; 
    double GetHeight()  return height; 
    char GetName()  return *name; 
;

int main()

    int n;
    City *mA;
    cout << "Input number of cities: " << endl;
    cin >> n;
    City *mA = new City[n];

    //read input
    for(int j = 0; j < n; j++)
        mA[j].Input();

    //bubble sort:
    for(int i = 0; i < n; i++)
    
        for(int j = i + 1; j < n; j++)
        
            if(mA[i].GetLength() > mA[j].GetLength())
            
                //swap values:
                City temp = mA[i];
                mA[i] = mA[j];
                mA[j] = temp;
            
        
    

    for(int j = 0; j < n; j++)
        mA[j].Output();

    delete[]mA;

    system("pause");
    return 0;

【参考方案2】：

手动排序需要比较重新排序的条目 - - - - 评论 - - - - - for (int j = 1; jmA[j].GetLength()) ... 交换元素 ... j-=2; //指数 ...如果（j

如果将使用 std::sort 条件是否可靠由 STL 检查。 并且只会检查回调函数的比较条件。

问候，休伯特·赫曼努茨