C++ 获取分隔符的迭代器位置

Question

谁能帮帮我，我的字符串迭代器有问题。

#include <iostream>
#include <string>
#include <vector>

using namespace std;

string s("foo $deli: baa :deli$ no matter")
string deli_open ("$deli:");
string deli_close (":deli$");

string::const_iterator it_begin = s.begin();
string::const_iterator it_end = s.begin();

// calculate won't work because I need the Iterposition! 
size_t found = s.find(deli_open);

if (found != string::npos)
   cout << found << '\n';

// your idea

// return $deli: baa :deli$
for( string::const_iterator i = it_begin; i != it_end; ++i)
{
  cout << (*i) ;
}

cout << endl;

这应该是解决方案:

返回 iteratorposion it_begin at "$deli: baa :deli$ no matter"

返回位于“foo $deli: baa :deli$”的迭代器位置 it_end

Answer 1

目前还不完全清楚你在问什么——我假设你正在寻找一个指向结束定界符末尾的迭代器。
这可以通过以下步骤完成:

1. 找到开始定界符后，从那里搜索结束定界符。

2. 找到后(从该搜索中获得的迭代器将指向它的开头)，将迭代器前进 N 个位置，N 是结束分隔符的长度:

   using namespace std;
   int main() {
     string s("foo $deli: baa :deli$ no matter")
     string deli_open ("$deli:");
     string deli_close (":deli$");
   
     size_t startOpenDelim = s.find(deli_open);
     if (startOpenDelim == string::npos)
       cerr << "opening delimiter not found\n";
   
     size_t startCloseDelim = s.find(deli_close, startOpenDelim);
     if (startCloseDelim == string::npos)
       cerr << "closing delimiter not found\n";
   
     size_t endCloseDelim = startCloseDelim + deli_close.length();
   
     string::const_iterator it_begin = s.begin() + startOpenDelim;
     string::const_iterator it_end = s.begin() + endCloseDelim;
   
     cout << string(it_begin, it_end) << endl;
   }