作业5-继承和派生
1.全面的MyString
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-
1 /*使程序输出指定结果*/ 2 #include <cstdlib> 3 #include <iostream> 4 using namespace std; 5 int strlen(const char * s) 6 { int i = 0; 7 for(; s[i]; ++i); 8 return i; 9 } 10 11 void strcpy(char * d,const char * s) 12 { 13 int i = 0; 14 for( i = 0; s[i]; ++i) 15 d[i] = s[i]; 16 d[i] = 0; 17 18 } 19 20 int strcmp(const char * s1,const char * s2) 21 { 22 for(int i = 0; s1[i] && s2[i] ; ++i) { 23 if( s1[i] < s2[i] ) 24 return -1; 25 else if( s1[i] > s2[i]) 26 return 1; 27 } 28 return 0; 29 } 30 31 void strcat(char * d,const char * s) 32 { 33 int len = strlen(d); 34 strcpy(d+len,s); 35 } 36 37 class MyString 38 { 39 // 在此处补充你的代码 40 char *p; 41 public: 42 MyString(const char * s = NULL) { //擅于利用缺省的参数，不要再写一个 43 if(s) { 44 p = new char[strlen(s) + 1]; 45 strcpy(p,s); 46 } 47 else 48 p = NULL; 49 50 } 51 ~MyString() { if(p) delete [] p; } 52 MyString(const MyString & s){ 53 if(s.p==NULL) p = NULL; 54 else{ 55 p = new char[strlen(s.p) + 1]; 56 strcpy(p,s.p); 57 } 58 } 59 MyString & operator=(const MyString & s) { 60 if(p) delete [] p; 61 if(s.p==NULL){ 62 p = NULL; return * this; 63 } 64 p = new char[strlen(s.p)+1]; 65 strcpy(p, s.p); 66 return * this; 67 } 68 MyString & operator=(const char * s) { 69 if(p) delete [] p; 70 if(s==NULL){ 71 p = NULL; return * this; 72 } 73 p = new char[strlen(s)+1]; 74 strcpy(p, s); 75 return * this; 76 } 77 char & operator[](const int i){ 78 return p[i]; 79 } 80 friend MyString operator+(const MyString & a, const MyString & b){ 81 MyString ans; 82 ans.p = new char[strlen(a.p)+strlen(b.p)+1]; 83 strcpy(ans.p, a.p); 84 strcat(ans.p, b.p); //不能改变原来的a 85 return ans; 86 } 87 MyString & operator+=(const char * s) { 88 MyString temp(s); 89 *this = *this+temp; 90 return * this; 91 } 92 bool operator<(const MyString s){ 93 return (strcmp(p, s.p)==-1); 94 } 95 bool operator>(const MyString s){ 96 return (strcmp(p, s.p)==1); 97 } 98 bool operator==(const MyString s){ 99 return (strcmp(p, s.p)==0); 100 } 101 char * operator()(int start, int len){ 102 char *temp = new char[len+1]; 103 for (int i = start; i < start+len; i++){ 104 temp[i-start] = p[i]; 105 } 106 temp[len] = '\0'; 107 return temp; 108 } 109 friend ostream & operator<<(ostream & o, const MyString & s){ 110 if(s.p==NULL )return o; // 空指针就不输出了 111 o<<s.p; 112 return o; 113 } 114 115 }; 116 117 118 int CompareString( const void * e1, const void * e2) 119 { 120 MyString * s1 = (MyString * ) e1; 121 MyString * s2 = (MyString * ) e2; 122 if( * s1 < *s2 ) //重载小于号 123 return -1; 124 else if( *s1 == *s2) 125 return 0; 126 else if( *s1 > *s2 ) //重载大于号 127 return 1; 128 } 129 130 int main() 131 { 132 MyString s1("abcd-"),s2,s3("efgh-"),s4(s1); //构造函数、复制构造函数 133 MyString SArray[4] = {"big","me","about","take"}; 134 cout << "1. " <<s1<<s2<<s3<< s4<< endl; //重载流插入 不能读 空指针 135 s4 = s3; //重载等于号 136 s3 = s1 + s3; // 重载+号 137 cout << "2. " << s1 << endl; 138 cout << "3. " << s2 << endl; 139 cout << "4. " << s3 << endl; 140 cout << "5. " << s4 << endl; 141 cout << "6. " << s1[2] << endl; 142 s2 = s1; 143 s1 = "ijkl-"; //重载等于号 144 s1[2] = 'A' ; // 重载中括号 145 cout << "7. " << s2 << endl; 146 cout << "8. " << s1 << endl; 147 s1 += "mnop"; //重载+= 148 cout << "9. " << s1 << endl; 149 s4 = "qrst-" + s2; 150 cout << "10. " << s4 << endl; 151 s1 = s2 + s4 + " uvw " + "xyz"; 152 cout << "11. " << s1 << endl; 153 qsort(SArray,4,sizeof(MyString),CompareString); 154 for( int i = 0;i < 4;i ++ ) 155 cout << SArray[i] << endl; 156 //s1的从下标0开始长度为4的子串 157 cout << s1(0,4) << endl; //重载小括号 158 //s1的从下标5开始长度为10的子串 159 cout << s1(5,10) << endl; 160 return 0; 161 }
备注：这道题其实没涉及到继承和派生，就是重温运算符重载。唯一一个有意思的地方，就是cout<<字符串空指针会导致后面的输出都没办法继续输出，所以在重载的时候要判断一下，如果是空指针就不输出才行。还有就是重温一下重载[]和（），注意一下返回值是什么。还有就是善于利用参数缺省的构造函数，这样就不用写两个构造函数了qwq
