5w5:第五周程序填空题1

描述

写一个MyString 类，使得下面程序的输出结果是：

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-

要求：MyString类必须是从C++的标准类string类派生而来。提示1：如果将程序中所有 "MyString" 用"string" 替换，那么题目的程序中除了最后两条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

程序：

#include <iostream>
#include <cstring>
#include <cstdlib>
using namespace std;

// 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;
    //输出s1从下标0开始长度为4的子串
    cout << s1(0,4) << endl;
    //输出s1从下标为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-

 

找了几个参考答案，但是提交的时候都失败了。

 

#include <iostream>
#include <string>
#include <cstdlib>
using namespace std;
class MyString : public string
{
public:
    MyString() {};
    //1.0继承类继承父类所有的成员变量和成员函数，但不继承构造函数和析构函数 
    //1.1继承类的无参构造函数，会隐式调用父类的无参构造函数 
    MyString(const char * st) :string(st) {};
    //1.2继承类的有参构造函数，如果父类也有有参构造函数，则必须显示调用它 
    //2.0这里的参数根据reference有两种选择，此处必须用const char*,"xxx"的类型是const char* 
    MyString(const MyString& s):string(s){}
    //1.3继承类的复制构造函数必须要显示的调用父类的复制构造函数，不然就会默认调用父类的无参构造函数 
    MyString operator +(MyString & op2)
    {
        string s1 = *this;
        string s2 = op2;
        string s = s1 + s2;
        return *new MyString(s.c_str());
    }
    MyString & operator +(const char * cs2)
    {
        string str1 = *this;
        string s = str1 + cs2;
        return *new MyString(s.c_str());
    }
    
    MyString & operator()(int s, int l)
    {
        string str = substr(s, l);
        return *new MyString(str.c_str());
    }
};

MyString operator+(const char * op1, MyString & op2)
{
    string st2 = op2;
    string s = op1 + st2;
    return *new MyString(s.c_str());
}

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-");
    MyString s4(s1);
    MyString SArray[4] = { "big","me","about","take" };
    //这里等号右边的赋值操作相当于调用了MyString的转换构造函数，其实就是单一非const classname&参数的构造函数可以直接接受参数类型的变量
    cout << "1. " << s1 << s2 << s3 << s4 << endl;
    s4 = s3;
    //3.0 operator=可以直接用string类里面的 
    s3 = s1 + 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;
    cout << s1(0, 4) << endl;
    cout << s1(5, 10) << endl;
    return 0;
}
--------------------- 
作者：qq_23908539 
来源：CSDN 
原文：https://blog.csdn.net/qq_23908539/article/details/51454521 
版权声明：本文为博主原创文章，转载请附上博文链接！

　　

这一个解释的还算比较详细，贴在上面供以后复习使用。