formatting a string

...and in answer to your deleted question

#include <iostream>
using namespace std;

int main(void)
{
     static     char strData[128];
     if(!gets(strData))
          {
          return 1;
          }
     short     numLength=strlen(strData);
     for(short numLoop=0; numLoop < numLength; numLoop++)
          {
          if ('=' == strData[numLoop] && ' ' ==strData[numLoop+1])
               {
               strData[numLoop]=0x0d;
               strData[numLoop+1]=0x0a;
               }
          }
     cout << strData;
     return 0;
}

Hi ,
U can use strtok to do that..

#include <string.h>
#include <stdio.h>

char string[] = "A string\tof ,,tokens\nand some  more tokens";
char seps[]   = " ,\t\n";
char *token;

void main( void )
{
   printf( "%s\n\nTokens:\n", string );
   /* Establish string and get the first token: */
   token = strtok( string, seps );
   while( token != NULL )
   {
      /* While there are tokens in "string" */
      printf( " %s\n", token );
      /* Get next token: */
      token = strtok( NULL, seps );
   }
}

Thanks for the solution.

Yes, akohli, you can use strtok(), but there are other things you have to do depending on your use of it.
If you want to leave it as one string with the CR/LF in it, you'll have to piece it back together.  There's nothing wrong with that.  In most cases, I would use strtok().  For this example, I took an easy route.

For a one to one replacement, a simpler method is ot use the replace_copy function.
Example:
#include <algorithm>

int main(int, char*)
{
     char data[] = "television+video+telephone+cassette recorder+";
     std::replace_copy(data,data+strlen(data),data,'+','\n');
     printf("%s\n", data);

     system("pause");
     return 0;
}

Correction, use the replace function, even simpler.

Example:
char data[] = "television+video+telephone+cassette recorder+";
std::replace(data,data+strlen(data), '+','\n');