splitting a string but keeping empty tokens c++

Question

I am trying to split a string and put it into a vector

however, I also want to keep an empty token whenever there are consecutive delimiter:

For example:

string mystring = "::aa;;bb;cc;;c"

I would like to tokenize this string on :; delimiters but in between delimiters such as :: and ;; I would like to push in my vector an empty string;

so my desired output for this string is:

"" (empty)
aa
"" (empty)
bb
cc
"" (empty)
c

Also my requirement is not to use the boost library.

if any could lend me an idea.

thanks

code that tokenize a string but does not include the empty tokens

void Tokenize(const string& str,vector<string>& tokens, const string& delim)
{
       // Skip delimiters at beginning.
     string::size_type lastPos = str.find_first_not_of(delimiters, 0);
     // Find first "non-delimiter".
     string::size_type pos     = str.find_first_of(delimiters, lastPos);

while (string::npos != pos || string::npos != lastPos)
 {
    // Found a token, add it to the vector.
    tokens.push_back(str.substr(lastPos, pos - lastPos));
    // Skip delimiters.  Note the "not_of"
    lastPos = str.find_first_not_of(delimiters, pos);
    // Find next "non-delimiter"
    pos = str.find_first_of(delimiters, lastPos);
  }
}

Answer 1

You can make your algorithm work with some simple changes. First, don't skip delimiters at the beginning, then instead of skipping delimiters in the middle of the string, just increment the position by one. Also, your npos check should ensure that both positions are not npos so it should be && instead of ||.

void Tokenize(const string& str,vector<string>& tokens, const string& delimiters)
{
    // Start at the beginning
    string::size_type lastPos = 0;
    // Find position of the first delimiter
    string::size_type pos = str.find_first_of(delimiters, lastPos);

    // While we still have string to read
    while (string::npos != pos && string::npos != lastPos)
    {
        // Found a token, add it to the vector
        tokens.push_back(str.substr(lastPos, pos - lastPos));
        // Look at the next token instead of skipping delimiters
        lastPos = pos+1;
        // Find the position of the next delimiter
        pos = str.find_first_of(delimiters, lastPos);
    }

    // Push the last token
    tokens.push_back(str.substr(lastPos, pos - lastPos));
}