【程设の旅】继承作业
感觉第一题没什么好说的 而且我觉得吧....
其实这个作业对继承的理解提升程度没有想象中的大
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-
Solution
输入vscode按照报错模拟即可
代码如下:
#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
{
private:
char *p;
public:
MyString(){
p=NULL;
}
MyString(const char *s){
p=new char[strlen(s)+1];
strcpy(p,s);
}
~MyString(){
if(p) delete[] p;
}
MyString(const MyString &s){
if(s.p==NULL) p=NULL;
else{
p=new char[strlen(s.p)+1];
strcpy(p,s.p);
}
}
MyString &operator=(const MyString &s){
if(p) delete[] p;
if(s.p==NULL){
p=NULL;
return *this;
}
else{
p=new char[strlen(s.p)+1];
strcpy(p,s.p);
return *this;
}
}
MyString &operator=(const char *s){
if(p) delete[] p;
if(s==NULL){
p=NULL;
return *this;
}
else{
p=new char[strlen(s)+1];
strcpy(p,s);
return *this;
}
}
char &operator[](int l){
return p[l];
}
friend MyString operator+(const MyString & a, const MyString & b){
MyString ans;
ans.p = new char[strlen(a.p)+strlen(b.p)+1];
strcpy(ans.p, a.p);
strcat(ans.p, b.p);
return ans;
}
MyString &operator+=(const char *c){
MyString temp(c);
*this=*this+temp;
return *this;
}
bool operator<(const MyString s){
return (strcmp(p,s.p)==-1);
}
bool operator==(const MyString s){
return (strcmp(p,s.p)==0);
}
bool operator>(const MyString s){
return (strcmp(p,s.p)==1);
}
char* operator()(int start,int len){
char *temp=new char[len+1];
for(int i=start;i<start+len;i++){
temp[i-start]=p[i];
}
temp[len]='\0';
return temp;
}
friend ostream &operator<<(ostream &o,const MyString &s){
if(s.p==NULL) return o;
else{
o<<s.p;
return o;
}
}
};
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;
system("pause");
return 0;
}
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
Solution
其实题目的意思 就是让我们写一个类 作为string的派生类
然后所有的数据都能用string的啦！
写三个构造函数 重载一个()即可
#include <cstdlib>
#include <iostream>
#include <string>
#include <algorithm>
using namespace std;
class MyString:public string
{
public:
MyString():string(){
}
MyString(const string s):string(s){
}
MyString(const char *s):string(s){}
string operator()(int start,int len){
return substr(start,len);
}
};
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;
system("pause");
return 0;
}
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