如何在c ++ 11 std :: regex中以其开始位置迭代std :: string中的所有正则表达式匹配项？ [英] how to iterate all regex matches in a std::string with their starting positions in c++11 std::regex?

问题描述

我知道两种从std :: string获取正则表达式匹配项的方法，但我不知道如何获取所有具有各自偏移量的匹配项。

I know two ways of getting regex matches from std::string, but I don't know how to get all matches with their respective offsets.

#include <string>
#include <iostream>
#include <regex>
int main() {
  using namespace std;
  string s = "123 apples 456 oranges 789 bananas oranges bananas";
  regex r = regex("[a-z]+");
  const sregex_token_iterator end;
  // here I know how to get all occurences
  // but don't know how to get starting offset of each one
  for (sregex_token_iterator i(s.cbegin(), s.cend(), r); i != end; ++i) {
    cout << *i << endl;
  }
  cout << "====" << endl;
  // and here I know how to get position
  // but the code is finding only first match
  smatch m;
  regex_search ( s, m, r );
  for (unsigned i=0; i< m.size(); ++i) {
    cout << m.position(i) << endl;
    cout << m[i] << endl;
  }
}

推荐答案

第一个首先，为什么要使用令牌迭代器？您没有要迭代的任何标记子表达式。

First of all, why token iterator? You don't have any marked subexpressions to iterate over.

第二， position（）是比赛的成员函数，因此：

Second, position() is a member function of a match, so:

#include <string>
#include <iostream>
#include <regex>
int main() {
    std::string s = "123 apples 456 oranges 789 bananas oranges bananas";
    std::regex r("[a-z]+");

    for(std::sregex_iterator i = std::sregex_iterator(s.begin(), s.end(), r);
                            i != std::sregex_iterator();
                            ++i )
    {
        std::smatch m = *i;
        std::cout << m.str() << " at position " << m.position() << '\n';
    }
}

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