北大程序设计与算法(三)测验题汇总(2020春季)
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
- abcd-efgh-abcd-
- abcd-
- abcd-efgh-
- efgh-
- c
- abcd-
- ijAl-
- ijAl-mnop
- qrst-abcd-
- abcd-qrst-abcd- uvw xyz
about
big
me
take
abcd
qrst-abcd-
样例输入
无
样例输出
- abcd-efgh-abcd-
- abcd-
- abcd-efgh-
- efgh-
- c
- abcd-
- ijAl-
- ijAl-mnop
- qrst-abcd-
- abcd-qrst-abcd- uvw xyz
about
big
me
take
abcd
qrst-abcd-
来源
Guo Wei
分析
首先定义一个成员,用于存储字符串
private:
char *str;
MyString s1("abcd-"),s2,s3("efgh-"),s4(s1);
构造函数构造;
MyString(){
str = new char[2];
str[0] = '\0';
}
MyString(const char *_str){//字符串
str = new char[strlen(_str) + 1];
strcpy(str,_str);
}
MyString(const MyString & s){//复制构造
if(strlen(str) < strlen(s.str)){
delete []str;
str = new char[strlen(s.str) + 1];
}
strcpy(str,s.str);
}
<<
的重载
cout << "1. " << s1 << s2 << s3<< s4<< endl;
friend ostream & operator << ( ostream & o,const MyString & s) {
o << s.str ;
return o;
}
=
的重载
s4 = s3;
MyString & operator=(const MyString & s) {
if(str == s.str)
return *this;
if(strlen(str) < strlen(s.str)){
delete []str;
str = new char[strlen(s.str) + 1];
}
strcpy(str,s.str);
return *this;
}
返回对象的引用目的是保证可以连等。
+
的重载
s3 = s1 + s3;
MyString operator + ( const MyString & s ) {
char *temp = new char[strlen(str) + strlen(s.str) + 2];
strcpy(temp,str);
strcat(temp,s.str);
MyString os(temp);
delete []temp;
return os;
}
注意此处不能是返回引用,因为如果返回引用的话相当于s1 + s3
生成的临时对象的引用传给s3显然这是不合理的,类型不匹配。
5. []
的重载
cout << "6. " << s1[2] << endl;
s1[2] = 'A' ;
char & operator[](int i){
return str[i];
}
返回参数的应用的目的是保证第二条赋值代码顺利进行。
6. =
重载
s1 = "ijkl-";
可能有同学会想,这儿难道不可以调用构造函数成临时对象在利用等号重载吗?其实,这儿如果调用构造函数就必须是初定义的时候才会调用,而s1显然不是此时初定义。
MyString & operator=(const char * s ) {
if(strlen(str) < strlen(s)){
delete []str;
str = new char[strlen(s) + 1];
}
strcpy(str,s);
return *this;
}
+
重载
s4 = "qrst-" + s2;
friend MyString operator +( const char * s1,const MyString & s2) {
MyString tmp(s1);
tmp+= s2;
return tmp;
}
+=
重载
s1 += "mnop";
//此处会发生临时对象创建
MyString & operator += ( const MyString & s) {
char *temp = new char[strlen(str) + strlen(s.str) + 2];
strcpy(temp,str);
strcat(temp,s.str);
delete []str;
str = temp;
return *this;
}
()
重载
MyString operator()(int start,int len) const {
char * tmp = new char[len + 1];
for( int i = 0;i < len ; ++i)
tmp[i] = str[start+i];
tmp[len] = 0;
MyString s(tmp);
delete [] tmp;
return s;
}
其中调用构造函数的有
MyString s1("abcd-"),s2,s3("efgh-"),s4(s1);
MyString SArray[4] = {"big","me","about","take"};//类型转换
s1 = "ijkl-";//类型转换
s1 += "mnop";//类型转换
s1 = s2 + s4 + " uvw " + "xyz";//类型转换
值得说明的是,在测试中发现,如果s4 = "cxasv" + s2;
是按照s4 = s2 + "cw";
的方式书写的话,其实直接调用类型转换构造函数即可,不需要额外重载+
。
至于s1 = s2 + s4 + " uvw " + "xyz";//类型转换
的计算顺序,我想了很久,经过测试,还是按照从左到右的顺序计算,此处利用+
重载。