如何迭代std::string中所有匹配的正则表达式，使其在c++11 std::regex中的起始位置

how to iterate all regex matches in a std::string with their starting positions in c++11 std::regex?

本文关键字：std 正则表达式 位置 c++11 regex 迭代 何迭代 string      更新时间：2023-10-16

我知道从std::string获得regex匹配的两种方法，但我不知道如何获得具有各自偏移量的所有匹配。

#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;
  }
}

首先，为什么是记号迭代器?没有任何标记的子表达式要迭代。

第二，position()是匹配的成员函数，所以: 
#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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