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 includecount   -   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";    }}