对字符串进行标记并在C ++中包含分隔符

Question

I'm tokening with the following, but unsure how to include the delimiters with it.

void Tokenize(const string str, vector<string>& tokens, const string& delimiters)
{

    int startpos = 0;
    int pos = str.find_first_of(delimiters, startpos);
    string strTemp;


    while (string::npos != pos || string::npos != startpos)
    {

        strTemp = str.substr(startpos, pos - startpos);
        tokens.push_back(strTemp.substr(0, strTemp.length()));

        startpos = str.find_first_not_of(delimiters, pos);
        pos = str.find_first_of(delimiters, startpos);

    }
}

Answer 1

The C++ String Toolkit Library (StrTk) has the following solution:

std::string str = "abc,123 xyz";
std::vector<std::string> token_list;
strtk::split(";., ",
             str,
             strtk::range_to_type_back_inserter(token_list),
             strtk::include_delimiters);

It should result with token_list have the following elements:

Token0 = "abc,"
Token1 = "123 "
Token2 = "xyz"

More examples can be found Here

Answer 2

I now this a little sloppy, but this is what I ended up with. I did not want to use boost since this is a school assignment and my instructor wanted me to use find_first_of to accomplish this.

Thanks for everyone's help.

vector<string> Tokenize(const string& strInput, const string& strDelims)
{
 vector<string> vS;

 string strOne = strInput;
 string delimiters = strDelims;

 int startpos = 0;
 int pos = strOne.find_first_of(delimiters, startpos);

 while (string::npos != pos || string::npos != startpos)
 {
  if(strOne.substr(startpos, pos - startpos) != "")
   vS.push_back(strOne.substr(startpos, pos - startpos));

  // if delimiter is a new line (\n) then addt new line
  if(strOne.substr(pos, 1) == "\n")
   vS.push_back("\\n");
  // else if the delimiter is not a space
  else if (strOne.substr(pos, 1) != " ")
   vS.push_back(strOne.substr(pos, 1));

  if( string::npos == strOne.find_first_not_of(delimiters, pos) )
   startpos = strOne.find_first_not_of(delimiters, pos);
  else
   startpos = pos + 1;

        pos = strOne.find_first_of(delimiters, startpos);

 }

 return vS;
}

Answer 3

I can't really follow your code, could you post a working program?

Anyway, this is a simple tokenizer, without testing edge cases:

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

using namespace std;

void tokenize(vector<string>& tokens, const string& text, const string& del)
{
    string::size_type startpos = 0,
        currentpos = text.find(del, startpos);

    do
    {
        tokens.push_back(text.substr(startpos, currentpos-startpos+del.size()));

        startpos = currentpos + del.size();
        currentpos = text.find(del, startpos);
    } while(currentpos != string::npos);

    tokens.push_back(text.substr(startpos, currentpos-startpos+del.size()));
}

Example input, delimiter = $$ :

Hello$$Stack$$Over$$$Flow$$$$!

Tokens:

Hello$$
Stack$$
Over$$
$Flow$$
$$
!

Note: I would never use a tokenizer I wrote without testing! please use boost::tokenizer !

Answer 4

如果分隔符是字符而不是字符串，那么你可以使用strtok 。

Answer 5

It depends on whether you want the preceding delimiters, the following delimiters, or both, and what you want to do with strings at the beginning and end of the string that may not have delimiters before/after them.

I'm going to assume you want each word, with its preceding and following delimiters, but NOT any strings of delimiters by themselves (eg if there's a delimiter following the last string).

template <class iter>
void tokenize(std::string const &str, std::string const &delims, iter out) { 
    int pos = 0;
    do { 
        int beg_word = str.find_first_not_of(delims, pos);
        if (beg_word == std::string::npos) 
            break;
        int end_word = str.find_first_of(delims, beg_word);
        int beg_next_word = str.find_first_not_of(delims, end_word);
        *out++ = std::string(str, pos, beg_next_word-pos);
        pos = end_word;
    } while (pos != std::string::npos);
}

For the moment, I've written it more like an STL algorithm, taking an iterator for its output instead of assuming it's always pushing onto a collection. Since it depends (for the moment) in the input being a string, it doesn't use iterators for the input.