string类型容器的相关操作（二）

本文通过一个程序实例来说明sting类型的相关操作。相关知识点见《C++ Primer》P293～297.

程序清单：

#include <iostream>
#include <string>

using namespace std;

int main(int argc,char *argv[])
{
    string s("Hello world!");

    string s1 = s.substr();
    cout << s1 << endl;

    string s2 = s.substr(6);
    cout << s2 << endl;

    string s3 = s.substr(6,5);
    cout << s3 << endl;

    string s4("C++ Primer");
    s4.append(" 3rd Ed."); //s4 == "C++ Primer 3rd Ed.";

    /*equivalent to s4.append(" 3rd Ed.")*/
// s4.insert(s4.size()," 3rd Ed.");

    cout << s4 << endl;

    s4.replace(11,3,"4th");
    /*equivalent way to replace "3rd" by "4th"*/
// s4.erase(11,3);

// s4.insert(11,"4th");

    cout << s4 << endl;

    /*不要求删除的文本长度与插入的相同*/
    s4.replace(11,3,"Fourth");
    cout << s4 << endl;

    string name("Yulingui Zhouyujia Gouyongpan");
    string::size_type pos1 = name.find("Zhouyujia");
    cout << pos1 << endl;

    /*find操作区分大小写*/
    string::size_type pos2 = name.find("zhouyujia");
    cout << pos2 << endl;
 
    /*查找任意字符*/
    string::size_type pos3 = name.find_first_of("iuoa");
    cout << "Find character at index: " << pos3
         << " element is "<< name[pos3] << endl;
 
    /*指定查找的起点*/
    string::size_type pos4 = 0;
    while((pos4=name.find_first_of("iuoa",pos4)) != string::npos)
    {
        cout << "Find character at index: " << pos4
             << " element is " << name[pos4] << endl;
        ++pos4;
    }

    /*寻找不匹配点*/
    /*在string对象中寻找第一个非数字字符*/
    string numbers("0123456789");
    string str("1596s8a99");
    string::size_type pos5 = str.find_first_not_of(numbers);
    cout << "The element at index " << pos5
         << " is the first element that is not a number."
         << " The element is " << str[pos5] << endl;
    
    return 0;
}

程序运行结果：

Hello world!

world!

world

C++ Primer 3rd Ed.

C++ Primer 4th Ed.

C++ Primer Fourth Ed.

9

4294967295

Find character at index: 1  element is u

Find character at index: 1  element is u

Find character at index: 3  element is i

Find character at index: 6  element is u

Find character at index: 7  element is i

Find character at index: 11  element is o

Find character at index: 12  element is u

Find character at index: 14  element is u

Find character at index: 16  element is i

Find character at index: 17  element is a

Find character at index: 20  element is o

Find character at index: 21  element is u

Find character at index: 23  element is o

Find character at index: 27  element is a

The element at index 4 is the first element that is not a number. The element is s