第八周项目4-String类的构造

写一个能处理字符串的类。

#include <iostream>
#include <Cstring>
using namespace std;
class String
{
public:
    String( );                         //默认构造函数
    String(const char *s);
    String(String &str);               //构造函数
    ~String();
    void display( );
    friend String operator + (String &s1,String &s2 );
    friend String operator - (String &s1,String &s2 );
private:
    char *p;
    int len;                                       //字符型指针,用于指向字符串
};
String::String()
{
    len = 0;
    p = NULL;
}
String::String(const char *s)
{
    len = strlen(s);
    p = new char[len+1];
    strcpy(p,s);
}

String::String(String &str)
{
    len = str.len;
    if(p!=NULL) delete []p;   //当发生在赋值等情形时,原对象可能已经存在,需释放原有空间
    p = new char[len+1];
    strcpy(p,str.p);
}

String::~String()
{
    if(!p) delete []p;
}

void String::display( )     //输出p所指向的字符串
{
    cout<<p<<endl;
}

String operator + (String &s1, String &s2 )
{
    String s;
    s.len = s1.len+s2.len;
    s.p = new char[s.len+1];   //原误写char(s.len+1)
    strcpy(s.p,s1.p);
    strcat(s.p,s2.p);
    return s;
}

String operator - (String &s1, String &s2 )
{
    String s;
    //c1为截去尾部空格的字符串
    char *c1=new char[s1.len+1];
    strcpy(c1,s1.p);
    int i=s1.len-1;
    while(i>=0&&c1[i]==' ') --i;
    c1[i+1]='\0';

    //c2为去除前导空格的字符串
    char *c2=new char[s2.len+1];
    strcpy(c2,s2.p);
    i=0;
    while(i<s2.len&&c2[i]==' ') ++i;
    int j=0;
    while(i<s2.len&&c2[i]!='\0')
    {
        c2[j]=c2[i];
        ++i;
        ++j;
    }
    c2[j]='\0';

    //将这两部分接起来
    s.len = strlen(c1)+strlen(c2);
    s.p = new char[s.len+1];     //原误写char(s.len+1)
    strcpy(s.p,c1);
    strcat(s.p,c2);
    delete c1;
    delete c2;
    return s;
}

int main( )
{
    String string1("Aloha  "), string2(" Mayuko");
    string1.display();
    string2.display();
    String string3;
    string3 = string1 + string2;
    string3.display();
    string3 = string1 - string2;
    string3.display();
    return 0;
}

运行结果:


@ Mayuko

posted @ 2015-05-14 20:13  麻麻麻麻鱼鱼  阅读(115)  评论(0编辑  收藏  举报