C++中substr函数的用法

substr用法

basic_string substr( size_type pos = 0, size_type count = npos ) const;

Returns a substring [pos, pos+count). If the requested substring extends past the end of the string, or if count == npos, the returned substring is [pos, size()).

Parameters

pos -   position of the first character to include
count   -   length of the substring

Return value

String containing the substring [pos, pos+count).

Exceptions

std::out_of_range if pos > size()

Complexity

Linear in count

#include <string>
#include <iostream>
using namespace std;
int main()
{
    string a = "0123456789abcdefghij";
    // count is npos, returns [pos, size())
    //从位置十个开始取到末尾
    string sub1 = a.substr(10);
    cout << sub1 << '\n';   //abcdefghij

    // 从位置5开始取，取三个
    string sub2 = a.substr(5, 3); 
    cout << sub2 << '\n';   //567

    // 取最后三个
    string sub4 = a.substr(a.size()-3, 50);
    cout << sub4 << '\n';   //hij

    try {
        // pos is out of bounds, throws
        //当所取位置超出了字符串长度，将抛出异常
        string sub5 = a.substr(a.size()+3, 50);
        cout << sub5 << '\n';
    } catch(const std::out_of_range& e) {
        cout << "pos exceeds string size\n";
    }
}