将'C'中的char数组拆分为CSV

Question

I need to split a char array into CSV's. Actually we can do the reverse of it using strtok() like:

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

int main ()
{
 char str[] ="This,a,sample,string.";
 char * pch;
 printf ("Splitting string \"%s\" into tokens:\n",str);
 pch = strtok (str,",");
 while (pch != NULL)
 {
   printf ("%s\n",pch);
   pch = strtok (NULL, ",");
 }
 return 0;
}

But in my case, there's an char array suppose char bits[1024]="abcdefghijklmn" . I need to get the output as a,b,c,d,e,f,g,h,i,j,k,m,n .

Is there any function or library to do this ie in terms of raw meaning, for every character it has to put a comma.

Answer 1

Just iterate over the string until you hit the end-of-string '\\0' character. Or use the length of the data in the array (which may be smaller than the array size) and use a simple for loop.

Answer 2

you can use this simple function from old basic :

// ............................................................. string word at

char * word_at(char *tString, int upTo, char *dilim) {

     int wcount;
     char *rString, *temp;
    temp= (char *) malloc(sizeof(char) * (strlen(tString)+1));
    strcpy(temp, tString);

    rString= strtok(temp, dilim);
    wcount=1;

    while (rString != NULL){
        if (wcount==upTo) { 
     return rString;
        }
        rString= strtok(NULL, dilim);
        wcount++;
    }
   return tString ;
  }

parameter : string , index and character delimiter return : word : ( char *)

Answer 3

This works for a null terminated string. But it will leave a dangling comma at the end.

void tokenise(char *s, char *d)
{
 while(*d++ = *s++) *d++ = ',';
}

If you know the length of the string already, you can pass that through. This will not leave a dangling comma.

void tokenise(char *s, char *d, int length)
{
    int i = 0;
    while((*d++ = *s++) && ((i++)<(length-1))) *d++ = ',';
}

In both examples, s is a pointer to the source string and d points to the output tokenised string. It is up to the calling code to ensure the buffer d points to is sufficiently large.

Answer 4

If you find easy to implement it, then this could help you to start

char* split_all( char arr[], char ch )
{
    char *new, *ptr;

    new = ptr = calloc( 1, 2*strlen( arr ) ); // FIXME : Error checks

    for( ; *(arr + 1) ; new++, arr++ )
    {
            *new = *arr;
            new++;
            *new = ch;
    }

    *new = *arr;
    return ptr;
}

You can re-use, optimize this for your requirement. Its a quick and dirty solution, feel free to fix it..