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";
}
}