regex_token_iterator<>有時會丟失匹配的子字串

正如這個問題中所建議的，我使用 regex_token_iterator<> 將所有匹配的子字串排成一行。但是代碼有時會錯過行中的第二個匹配子字串，并且發生這種錯過的行會在不同的運行中發生變化。這是 regex_token_iterator<> 的錯誤，還是我的代碼有問題？我使用的編譯器是Apple clang version 14.0.0 (clang-1400.0.29.202)，我使用-std=c 14編譯了下面的代碼。

我在上面的問題中還嘗試了另一個建議，即使用while-loop重復應用regex_search()，并且該版本的代碼可以正常運行。我只想知道為什么帶有 regex_token_iterator<> 的版本不起作用，我的用法是否錯誤。

代碼：

#include<regex>
#include<iostream>
#include<string>
#include<fstream>
#include<sstream>

using namespace std;

struct bad_from_string : bad_cast{
  const char* what() const noexcept override{
    return "bad cast from string";
  }
};

template<typename T>
T from_string(const string& s){
  istringstream is{s};
  T t;
  if(!(is>>t))
    throw bad_from_string{};
  return t;
}

int main(){
  regex pat{R"((\d{1,2})/(\d{1,2})/(\d{4}))"}; // e.g. 7/21/2022
  ifstream ifs{"test_regex_token_iterator.txt"};
  ofstream ofs{"test_out_regex_token_iterator.txt"};

  regex_token_iterator<string::iterator> rend; // default constructor is used for indicating the end of the sequence
  
  for(string line; getline(ifs, line);){
    smatch matches;
    
    string replace_pattern; 

    int month{0}, day{0}, year{0};

    regex_token_iterator<string::iterator> riter(line.begin(), line.end(), pat);
      
    // for each matched substring, replace it individually
    while(riter!=rend){
      string matched_substring{(*riter).str()};
      // *riter returns a reference to the sub_match object riter is pointing to.
      // sub_match is not a string. sub_match::str() returns the string of the sub_match.
      
      // put each matched substring into variable "matches"
      regex_search(matched_substring, matches, pat);
      
      // get the day, month, and year values in int
      day = from_string<int>(matches.str(2));
      month = from_string<int>(matches.str(1));
      year = from_string<int>(matches.str(3));
      
      // here make replace_pattern yyyy-mm-dd
      if(month<10 && day<10)
        replace_pattern = to_string(year) "-0" to_string(month) "-0" to_string(day); // both day and month need the fron '0'
      else if(month<10)
        replace_pattern = to_string(year) "-0" to_string(month) "-" to_string(day);
      else if(day<10)
        replace_pattern = to_string(year) "-" to_string(month) "-0" to_string(day);
      else
        replace_pattern = to_string(year) "-" to_string(month) "-" to_string(day);
      
      line = regex_replace(line, regex(matched_substring), replace_pattern); // regex_replace() returns a string
      // since I want to replace only 1 matched substring *riter, I use the exact substring 
      // in the place of regex pattern
      
        riter;      // move to the next matched substring
    }
  
    ofs << line << endl; 
  }
  
  return 0;
}

test_regex_token_iterator.txt：

12/01/2022 - 12/31/2022
12/01/2022 - 12/31/2022
12/01/2022 - 12/31/2022
12/01/2022 - 12/31/2022

10/01/2022 - 10/31/2022
10/01/2022 - 10/31/2022
10/01/2022 - 10/31/2022
10/01/2022 - 10/31/2022
10/01/2022 - 10/31/2022

示例 test_out_regex_token_iterator.txt（但結果在不同的運行中會發生變化）：

2022-12-01 - 12/31/2022
2022-12-01 - 2022-12-31
2022-12-01 - 12/31/2022
2022-12-01 - 12/31/2022

2022-10-01 - 10/31/2022
2022-10-01 - 2022-10-31
2022-10-01 - 10/31/2022
2022-10-01 - 10/31/2022
2022-10-01 - 10/31/2022

我希望所有匹配的子字串（包括第二列中的日期）都被替換，但只有一部分被正確替換。預期結果：

2022-12-01 - 2022-12-31
2022-12-01 - 2022-12-31
2022-12-01 - 2022-12-31
2022-12-01 - 2022-12-31

2022-10-01 - 2022-10-31
2022-10-01 - 2022-10-31
2022-10-01 - 2022-10-31
2022-10-01 - 2022-10-31
2022-10-01 - 2022-10-31

uj5u.com熱心網友回復：

啟用地址消毒器表明您的代碼導致了未定義的行為：https ://godbolt.org/z/n3rnn9nqY

riter包含來自的迭代器，line但在 while 回圈的末尾，您重新分配line、使line迭代器無效并因此使迭代器無效riter，然后當您嘗試遞增時，riter您進入了未定義行為的領域。

為您的輸出添加一個單獨的字串可以解決問題：https ://godbolt.org/z/Grqe1vv5x

for(string line; getline(ifs, line);){
  smatch matches;
  string outputLine = line;
  
  string replace_pattern; 

  int month{0}, day{0}, year{0};

  regex_token_iterator<string::iterator> riter(line.begin(), line.end(), pat);
    
  // for each matched substring, replace it individually
  while(riter!=rend){
    string matched_substring{(*riter).str()};
    // *riter returns a reference to the sub_match object riter is pointing to.
    // sub_match is not a string. sub_match::str() returns the string of the sub_match.
    
    // put each matched substring into variable "matches"
    regex_search(matched_substring, matches, pat);
    
    // get the day, month, and year values in int
    day = from_string<int>(matches.str(2));
    month = from_string<int>(matches.str(1));
    year = from_string<int>(matches.str(3));
    
    // here make replace_pattern yyyy-mm-dd
    if(month<10 && day<10)
      replace_pattern = to_string(year) "-0" to_string(month) "-0" to_string(day); // both day and month need the fron '0'
    else if(month<10)
      replace_pattern = to_string(year) "-0" to_string(month) "-" to_string(day);
    else if(day<10)
      replace_pattern = to_string(year) "-" to_string(month) "-0" to_string(day);
    else
      replace_pattern = to_string(year) "-" to_string(month) "-" to_string(day);
    
    outputLine = regex_replace(outputLine, regex(matched_substring), replace_pattern); // regex_replace() returns a string
    // since I want to replace only 1 matched substring *riter, I use the exact substring 
    // in the place of regex pattern
    
      riter;      // move to the next matched substring
  }

  ofs << outputLine << endl; 
}

標籤：C 正则表达式

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