18.03.21 继承作业

A:全面的MyString

描述

程序填空，输出指定结果

#include <cstdlib>
#include <iostream>
using namespace std;
int strlen(const char * s) 
{    int i = 0;
    for(; s[i]; ++i);
    return i;
}
void strcpy(char * d,const char * s)
{
    int i = 0;
    for( i = 0; s[i]; ++i)
        d[i] = s[i];
    d[i] = 0;
        
}
int strcmp(const char * s1,const char * s2)
{
    for(int i = 0; s1[i] && s2[i] ; ++i) {
        if( s1[i] < s2[i] )
            return -1;
        else if( s1[i] > s2[i])
            return 1;
    }
    return 0;
}
void strcat(char * d,const char * s)
{
    int len = strlen(d);
    strcpy(d+len,s);
}
class MyString
{
// 在此处补充你的代码
};


int CompareString( const void * e1, const void * e2)
{
    MyString * s1 = (MyString * ) e1;
    MyString * s2 = (MyString * ) e2;
    if( * s1 < *s2 )
    return -1;
    else if( *s1 == *s2)
    return 0;
    else if( *s1 > *s2 )
    return 1;
}
int main()
{
    MyString s1("abcd-"),s2,s3("efgh-"),s4(s1);
    MyString SArray[4] = {"big","me","about","take"};
    cout << "1. " << s1 << s2 << s3<< s4<< endl;
    s4 = s3;
    s3 = s1 + s3;
    cout << "2. " << s1 << endl;
    cout << "3. " << s2 << endl;
    cout << "4. " << s3 << endl;
    cout << "5. " << s4 << endl;
    cout << "6. " << s1[2] << endl;
    s2 = s1;
    s1 = "ijkl-";
    s1[2] = 'A' ;
    cout << "7. " << s2 << endl;
    cout << "8. " << s1 << endl;
    s1 += "mnop";
    cout << "9. " << s1 << endl;
    s4 = "qrst-" + s2;
    cout << "10. " << s4 << endl;
    s1 = s2 + s4 + " uvw " + "xyz";
    cout << "11. " << s1 << endl;
    qsort(SArray,4,sizeof(MyString),CompareString);
    for( int i = 0;i < 4;i ++ )
    cout << SArray[i] << endl;
    //s1的从下标0开始长度为4的子串
    cout << s1(0,4) << endl;
    //s1的从下标5开始长度为10的子串
    cout << s1(5,10) << endl;
    return 0;
}

输入

无

输出

1. abcd-efgh-abcd-
2. abcd-
3. 
4. abcd-efgh-
5. efgh-
6. c
7. abcd-
8. ijAl-
9. ijAl-mnop
10. qrst-abcd-
11. abcd-qrst-abcd- uvw xyz
about
big
me
take
abcd
qrst-abcd-样例输入

无

样例输出

1. abcd-efgh-abcd-
2. abcd-
3. 
4. abcd-efgh-
5. efgh-
6. c
7. abcd-
8. ijAl-
9. ijAl-mnop
10. qrst-abcd-
11. abcd-qrst-abcd- uvw xyz
about
big
me
take
abcd
qrst-abcd-

来源Guo Wei

 1     char *p;
 2 public:
 3     //构造函数
 4     MyString(const char *a){
 5         //if(p)delete[]p;
 6         p=new char[strlen(a)+1];
 7         strcpy(p,a);
 8     }
 9     MyString(){
10         p=new char[1];
11         p[0]='';
12     }
13     //copy
14     MyString(const MyString&s){
15         p=new char[strlen(s.p)+1];
16         strcpy(p,s.p);
17     }
18     //destruct
19     ~MyString(){if(p)delete[]p;}
20     //cout
21     friend ostream&operator<<(ostream&os,MyString &a){
22         os<<a.p;
23         return os;
24     }
25     //class+class
26     friend MyString operator+(const MyString &a,const MyString &b){
27         char *res=new char[strlen(b.p)+strlen(a.p)+1];
28         strcpy(res,a.p);
29         strcat(res,b.p);
30         return MyString(res);
31     }
32     //[]
33     char&operator[](int x){
34         return p[x];
35     }
36     //class=class
37     MyString& operator=(const MyString&a){
38         if(p)delete[]p;
39         p=new char[strlen(a.p)+1];
40         strcpy(p,a.p);
41         return *this;
42     }
43     //class=char
44     MyString& operator=(char *s){
45         if(p)delete[]p;
46         p=new char[strlen(s)+1];
47         strcpy(p,s);
48         return *this;
49     }
50     //class+=char
51     MyString operator +=(const char *s){
52         *this=*this+MyString(s);
53         return *this;
54     }
55     //char+class
56     friend MyString operator +(char *s,MyString&a){
57         return MyString(s)+a;
58     }
59     //class+char
60     MyString operator +(char*s){
61         return *this+MyString(s);
62     }
63     //<
64     friend bool operator<(MyString&a,MyString&b){
65         return (strcmp(a.p,b.p)==-1);
66     }
67     //>
68     friend bool operator>(MyString&a,MyString&b){
69         return (strcmp(a.p,b.p)==1);
70     }
71     //=
72     friend bool operator==(MyString&a,MyString&b){
73         return (strcmp(a.p,b.p)==0);
74     }
75     //()
76     char*operator()(int x,int y){
77        char *ptr=new char[y+1];
78        for(int i=0;i<y;i++){
79            ptr[i]=p[i+x];
80        }
81        ptr[y]='';
82        return ptr;
83     }
View Code

去掉了if(p)delete[]p;一行就过了？？emmm……不知道是什么情况

以及vs里不好调……空手调bug

加个分析

 1 int main()
 2 {
 3     MyString s1("abcd-"),s2,s3("efgh-"),s4(s1);//构造函数和复制函数
 4     MyString SArray[4] = {"big","me","about","take"};
 5     cout << "1. " << s1 << s2 << s3<< s4<< endl;//输出流重载
 6     s4 = s3;
 7     s3 = s1 + s3;//加号重载 连接字符串
 8     cout << "2. " << s1 << endl;
 9     cout << "3. " << s2 << endl;
10     cout << "4. " << s3 << endl;
11     cout << "5. " << s4 << endl;
12     cout << "6. " << s1[2] << endl;//重构[] 返回引用
13     s2 = s1;//重载赋值
14     s1 = "ijkl-";//不知可否用构造函数直接解决
15     s1[2] = 'A' ;
16     cout << "7. " << s2 << endl;
17     cout << "8. " << s1 << endl;
18     s1 += "mnop";//重构+=
19     cout << "9. " << s1 << endl;
20     s4 = "qrst-" + s2;//全局函数重载+
21     cout << "10. " << s4 << endl;
22     s1 = s2 + s4 + " uvw " + "xyz";//又一个全局函数重载+ 并且重载都返MyString& 不对 临时不能返&
23     cout << "11. " << s1 << endl;
24     qsort(SArray,4,sizeof(MyString),CompareString);//重载<>==
25     for( int i = 0;i < 4;i ++ )
26     cout << SArray[i] << endl;
27     //s1的从下标0开始长度为4的子串 重载（）
28     cout << s1(0,4) << endl;
29     //s1的从下标5开始长度为10的子串
30     cout << s1(5,10) << endl;
31     return 0;
32 }
View Code

B:继承自string的MyString

描述

程序填空，输出指定结果

#include <cstdlib>
#include <iostream>
#include <string>
#include <algorithm>
using namespace std;
class MyString:public string
{
// 在此处补充你的代码
};


int main()
{
    MyString s1("abcd-"),s2,s3("efgh-"),s4(s1);
    MyString SArray[4] = {"big","me","about","take"};
    cout << "1. " << s1 << s2 << s3<< s4<< endl;
    s4 = s3;
    s3 = s1 + s3;
    cout << "2. " << s1 << endl;
    cout << "3. " << s2 << endl;
    cout << "4. " << s3 << endl;
    cout << "5. " << s4 << endl;
    cout << "6. " << s1[2] << endl;
    s2 = s1;
    s1 = "ijkl-";
    s1[2] = 'A' ;
    cout << "7. " << s2 << endl;
    cout << "8. " << s1 << endl;
    s1 += "mnop";
    cout << "9. " << s1 << endl;
    s4 = "qrst-" + s2;
    cout << "10. " << s4 << endl;
    s1 = s2 + s4 + " uvw " + "xyz";
    cout << "11. " << s1 << endl;
        sort(SArray,SArray+4);
    for( int i = 0;i < 4;i ++ )
    cout << SArray[i] << endl;
    //s1的从下标0开始长度为4的子串
    cout << s1(0,4) << endl;
    //s1的从下标5开始长度为10的子串
    cout << s1(5,10) << endl;
    return 0;
}

输入

无

输出

1. abcd-efgh-abcd-
2. abcd-
3. 
4. abcd-efgh-
5. efgh-
6. c
7. abcd-
8. ijAl-
9. ijAl-mnop
10. qrst-abcd-
11. abcd-qrst-abcd- uvw xyz
about
big
me
take
abcd
qrst-abcd-样例输入

无

样例输出

1. abcd-efgh-abcd-
2. abcd-
3. 
4. abcd-efgh-
5. efgh-
6. c
7. abcd-
8. ijAl-
9. ijAl-mnop
10. qrst-abcd-
11. abcd-qrst-abcd- uvw xyz
about
big
me
take
abcd
qrst-abcd-

提示提示 1：如果将程序中所有 "MyString" 用 "string" 替换，那么除
了最后两条红色的语句编译无法通过外，其他语句都没有问题，而且输出和前
面给的结果吻合。也就是说，MyString 类对 string 类的功能扩充只体现在最
后两条语句上面。 

提示 2: string 类有一个成员函数 string substr(int start,int 
length); 能够求从 start 位置开始，长度为 length 的子串 

提示 3: C++中，派生类的对象可以赋值给基类对象，因为，一个派生
类对象，也可看作是一个基类对象（大学生是学生）。反过来则不行(学生未
必是大学生） 同样，调用需要基类对象作参数的函数时，以派生类对象作为实参，也是没有问题的来源Guo Wei

 1 public:
 2     //construct
 3     MyString():string(""){}
 4     MyString(const char*s):string(s){}
 5     MyString(const MyString&a):string(a){}
 6     //+
 7     MyString operator+(const char *b){
 8         string str=*this;
 9         str+=b;
10         return MyString(str.c_str());
11     }
12     MyString operator+(const MyString&a){
13         string str1=*this,str2=a;
14         str1+=a;
15         return MyString(str1.c_str());
16     }
17 
18     friend MyString operator+(const char*s,const MyString &a){
19         string str1=a,str2=s;
20         str2+=str1;
21         return MyString(str2.c_str());
22     }
23     //+=
24     MyString& operator+=(const char*s){
25         *this=*this+s;
26         return *this;
27     }
28     //()
29     string operator()(int x,int y){
30         string str;
31         for(int i=0;i<y;i++)
32             str+=(*this)[x+i];
33         return MyString(str.c_str());
34     }
View Code

早点看到提示就好了……

注定失败的战争，也要拼尽全力去打赢它； 就算输，也要输得足够漂亮。
【推广】 免费学中医，健康全家人
原文地址：https://www.cnblogs.com/yalphait/p/8618195.html