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STL Split String

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1 Feb 2006 4  
A function that will split an input string based on a string delimiter.

Description

Below is a function I created and have found extremely useful for splitting strings based on a particular delimiter. The implementation only requires STL which makes it easy to port to any OS that supports STL. The function is fairly lightweight although I haven't done extensive performance testing.

The delimiter can be n number of characters represented as a string. The parts of the string in between the delimiter are then put into a string vector. The class StringUtils contains one static function SplitString. The int returned is the number of delimiters found within the input string.

I used this utility mainly for parsing strings that were being passed across platform boundaries. Whether you are using raw sockets or middleware such as TIBCO� it is uncomplicated to pass string data. I found it more efficient to pass delimited string data verses repeated calls or messages. Another place I used this was in passing BSTRs back and forth between a Visual Basic client and an ATL COM DLL. It proved to be easier than passing a SAFEARRAY as an [in] or [out] parameter. This was also beneficial when I did not want the added overhead of MFC and hence could not use CString.

Implementation

The SplitString function uses the STL string functions find and substr to iterate through the input string. The hardest part was figuring out how to get the substring of the input string based on the offsets of the delimiter, not forgetting to take into account the length of the delimiter. Another hurdle was making sure not to call substr with an offset greater than the length of the input string.

Header

#ifndef __STRINGUTILS_H_
#define __STRINGUTILS_H_

#include <string>

#include <vector>


using namespace std;

class StringUtils
{

public:

    static int SplitString(const string& input, 
        const string& delimiter, vector<string>& results, 
        bool includeEmpties = true);

};

#endif

Source

int StringUtils::SplitString(const string& input, 
       const string& delimiter, vector<string>& results, 
       bool includeEmpties)
{
    int iPos = 0;
    int newPos = -1;
    int sizeS2 = (int)delimiter.size();
    int isize = (int)input.size();

    if( 
        ( isize == 0 )
        ||
        ( sizeS2 == 0 )
    )
    {
        return 0;
    }

    vector<int> positions;

    newPos = input.find (delimiter, 0);

    if( newPos < 0 )
    { 
        return 0; 
    }

    int numFound = 0;

    while( newPos >= iPos )
    {
        numFound++;
        positions.push_back(newPos);
        iPos = newPos;
        newPos = input.find (delimiter, iPos+sizeS2);
    }

    if( numFound == 0 )
    {
        return 0;
    }

    for( int i=0; i <= (int)positions.size(); ++i )
    {
        string s("");
        if( i == 0 ) 
        { 
            s = input.substr( i, positions[i] ); 
        }
        int offset = positions[i-1] + sizeS2;
        if( offset < isize )
        {
            if( i == positions.size() )
            {
                s = input.substr(offset);
            }
            else if( i > 0 )
            {
                s = input.substr( positions[i-1] + sizeS2, 
                      positions[i] - positions[i-1] - sizeS2 );
            }
        }
        if( includeEmpties || ( s.size() > 0 ) )
        {
            results.push_back(s);
        }
    }
    return numFound;
}

Output using demo project

main.exe "|mary|had|a||little|lamb||" "|"

int SplitString(
        const string& input,
        const string& delimiter,
        vector<string>& results,
        bool includeEmpties = true
)

-------------------------------------------------------
input           = |mary|had|a||little|lamb||
delimiter       = |
return value    = 8 // Number of delimiters found

results.size()  = 9
results[0]      = ''
results[1]      = 'mary'
results[2]      = 'had'
results[3]      = 'a'
results[4]      = ''
results[5]      = 'little'
results[6]      = 'lamb'
results[7]      = ''
results[8]      = ''

int SplitString(
        const string& input,
        const string& delimiter,
        vector<string>& results,
        bool includeEmpties = false
)

-------------------------------------------------------
input           = |mary|had|a||little|lamb||
delimiter       = |
return value    = 8 // Number of delimiters found

results.size()  = 5
results[0]      = 'mary'
results[1]      = 'had'
results[2]      = 'a'
results[3]      = 'little'
results[4]      = 'lamb'

MFC version

For those of you who absolutely cannot use STL and are committed to MFC I made a few minor changes to the above implementation. It uses CString instead of std::string and a CStringArray instead of a std::vector:

//------------------------

// SplitString in MFC

//------------------------

int StringUtils::SplitString(const CString& input, 
  const CString& delimiter, CStringArray& results)
{
  int iPos = 0;
  int newPos = -1;
  int sizeS2 = delimiter.GetLength();
  int isize = input.GetLength();

  CArray<INT, int> positions;

  newPos = input.Find (delimiter, 0);

  if( newPos < 0 ) { return 0; }

  int numFound = 0;

  while( newPos > iPos )
  {
    numFound++;
    positions.Add(newPos);
    iPos = newPos;
    newPos = input.Find (delimiter, iPos+sizeS2+1);
  }

  for( int i=0; i <= positions.GetSize(); i++ )
  {
    CString s;
    if( i == 0 )
      s = input.Mid( i, positions[i] );
    else
    {
      int offset = positions[i-1] + sizeS2;
      if( offset < isize )
      {
        if( i == positions.GetSize() )
          s = input.Mid(offset);
        else if( i > 0 )
          s = input.Mid( positions[i-1] + sizeS2, 
                 positions[i] - positions[i-1] - sizeS2 );
      }
    }
    if( s.GetLength() > 0 )
      results.Add(s);
  }
  return numFound;
}

String neutral version

I added this version in case you might need to use it with any type of string. The only requirement is the string class must have a constructor that takes a char*. The code only depends on the STL vector. I also added the option to not include empty strings in the results, which will occur if delimiters are adjacent:

//-----------------------------------------------------------

// StrT:    Type of string to be constructed

//          Must have char* ctor.

// str:     String to be parsed.

// delim:   Pointer to delimiter.

// results: Vector of StrT for strings between delimiter.

// empties: Include empty strings in the results. 

//-----------------------------------------------------------

template< typename StrT >
int split(const char* str, const char* delim, 
     vector<StrT>& results, bool empties = true)
{
  char* pstr = const_cast<char*>(str);
  char* r = NULL;
  r = strstr(pstr, delim);
  int dlen = strlen(delim);
  while( r != NULL )
  {
    char* cp = new char[(r-pstr)+1];
    memcpy(cp, pstr, (r-pstr));
    cp[(r-pstr)] = '\0';
    if( strlen(cp) > 0 || empties )
    {
      StrT s(cp);
      results.push_back(s);
    }
    delete[] cp;
    pstr = r + dlen;
    r = strstr(pstr, delim);
  }
  if( strlen(pstr) > 0 || empties )
  {
    results.push_back(StrT(pstr));
  }
  return results.size();
}

String neutral usage

// using CString

//------------------------------------------

int i = 0;
vector<CString> results;
split("a-b-c--d-e-", "-", results);
for( i=0; i < results.size(); ++i )
{
  cout << results[i].GetBuffer(0) << endl;
  results[i].ReleaseBuffer();
}

// using std::string

//------------------------------------------

vector<string> stdResults;
split("a-b-c--d-e-", "-", stdResults);
for( i=0; i < stdResults.size(); ++i )
{
  cout << stdResults[i].c_str() << endl;
}

// using std::string without empties

//------------------------------------------

stdResults.clear();
split("a-b-c--d-e-", "-", stdResults, false);
for( i=0; i < stdResults.size(); ++i )
{
  cout << stdResults[i].c_str() << endl;
}

Conclusion

Hope you find this as useful as I did. Feel free to let me know of any bugs or enhancements. Enjoy ;)

License

This article has no explicit license attached to it but may contain usage terms in the article text or the download files themselves. If in doubt please contact the author via the discussion board below.

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