Split string with delimiters in C

Question

How do I write a function to split and return an array for a string with delimiters in the C programming language?

char* str = \"JAN,FEB,MAR,APR,MAY,JUN,JUL,         


        

Answer 1

Explode & implode - initial string remains intact, dynamic memory allocation

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

typedef struct
{
    uintptr_t   ptr;
    int         size;
} token_t;

int explode(char *str, int slen, const char *delimiter, token_t **tokens)
{
    int i = 0, c1 = 0, c2 = 0;

    for(i = 0; i <= slen; i++)
    {
            if(str[i] == *delimiter)
            {
                c1++;
            }
    }

    if(c1 == 0)
    {
            return -1;
    }

    *tokens = (token_t*)calloc((c1 + 1), sizeof(token_t));
    ((*tokens)[c2]).ptr = (uintptr_t)str;

    i = 0; 
    while(i <= slen)
    {
        if((str[i] == *delimiter) || (i == slen))
        {
                ((*tokens)[c2]).size = (int)((uintptr_t)&(str[i]) - (uintptr_t)(((*tokens)[c2]).ptr));
                if(i < slen)
                {
                    c2++;
                    ((*tokens)[c2]).ptr = (uintptr_t)&(str[i + 1]);
                }
        }
        i++;
    }
    return (c1 + 1);
}

char* implode(token_t *tokens, int size, const char *delimiter)
{
    int     i, len = 0;
    char    *str;

    for(i = 0; i < len; i++)
    {
        len += tokens[i].size + 1;
    }

    str = (char*)calloc(len, sizeof(char));

    len = 0;
    for(i = 0; i < size; i++)
    {
        memcpy((void*)&str[len], (void*)tokens[i].ptr, tokens[i].size);
        len += tokens[i].size;
        str[(len++)] = *delimiter;
    }

    str[len - 1] = '\0';

    return str;
}

Usage:

int main(int argc, char **argv)
{
    int         i, c;
    char        *exp = "JAN,FEB,MAR,APR,MAY,JUN,JUL,AUG,SEP,OCT,NOV,DEC";
    token_t     *tokens;
    char        *imp;

    printf("%s\n", exp);

    if((c = explode(exp, strlen(exp), ",", &tokens)) > 0)
    {
        imp = implode(tokens, c, ",");
        printf("%s\n", imp);

        for(i = 0; i < c; i++)
        {
            printf("%.*s, %d\n", tokens[i].size, (char*)tokens[i].ptr, tokens[i].size);
        }
    }

    free((void*)tokens);
    free((void*)imp);
    return 0;
}

Answer 2

In the above example, there would be a way to return an array of null terminated strings (like you want) in place in the string. It would not make it possible to pass a literal string though, as it would have to be modified by the function:

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

char** str_split( char* str, char delim, int* numSplits )
{
    char** ret;
    int retLen;
    char* c;

    if ( ( str == NULL ) ||
        ( delim == '\0' ) )
    {
        /* Either of those will cause problems */
        ret = NULL;
        retLen = -1;
    }
    else
    {
        retLen = 0;
        c = str;

        /* Pre-calculate number of elements */
        do
        {
            if ( *c == delim )
            {
                retLen++;
            }

            c++;
        } while ( *c != '\0' );

        ret = malloc( ( retLen + 1 ) * sizeof( *ret ) );
        ret[retLen] = NULL;

        c = str;
        retLen = 1;
        ret[0] = str;

        do
        {
            if ( *c == delim )
            {
                ret[retLen++] = &c[1];
                *c = '\0';
            }

            c++;
        } while ( *c != '\0' );
    }

    if ( numSplits != NULL )
    {
        *numSplits = retLen;
    }

    return ret;
}

int main( int argc, char* argv[] )
{
    const char* str = "JAN,FEB,MAR,APR,MAY,JUN,JUL,AUG,SEP,OCT,NOV,DEC";

    char* strCpy;
    char** split;
    int num;
    int i;

    strCpy = malloc( strlen( str ) * sizeof( *strCpy ) );
    strcpy( strCpy, str );

    split = str_split( strCpy, ',', &num );

    if ( split == NULL )
    {
        puts( "str_split returned NULL" );
    }
    else
    {
        printf( "%i Results: \n", num );

        for ( i = 0; i < num; i++ )
        {
            puts( split[i] );
        }
    }

    free( split );
    free( strCpy );

    return 0;
}

There is probably a neater way to do it, but you get the idea.

Answer 3

Try use this.

char** strsplit(char* str, const char* delim){
    char** res = NULL;
    char*  part;
    int i = 0;

    char* aux = strdup(str);

    part = strdup(strtok(aux, delim));

    while(part){
        res = (char**)realloc(res, (i + 1) * sizeof(char*));
        *(res + i) = strdup(part);

        part = strdup(strtok(NULL, delim));
        i++;
    }

    res = (char**)realloc(res, i * sizeof(char*));
    *(res + i) = NULL;

    return res;
}

Answer 4

You can use the strtok() function to split a string (and specify the delimiter to use). Note that strtok() will modify the string passed into it. If the original string is required elsewhere make a copy of it and pass the copy to strtok().

EDIT:

Example (note it does not handle consecutive delimiters, "JAN,,,FEB,MAR" for example):

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

char** str_split(char* a_str, const char a_delim)
{
    char** result    = 0;
    size_t count     = 0;
    char* tmp        = a_str;
    char* last_comma = 0;
    char delim[2];
    delim[0] = a_delim;
    delim[1] = 0;

    /* Count how many elements will be extracted. */
    while (*tmp)
    {
        if (a_delim == *tmp)
        {
            count++;
            last_comma = tmp;
        }
        tmp++;
    }

    /* Add space for trailing token. */
    count += last_comma < (a_str + strlen(a_str) - 1);

    /* Add space for terminating null string so caller
       knows where the list of returned strings ends. */
    count++;

    result = malloc(sizeof(char*) * count);

    if (result)
    {
        size_t idx  = 0;
        char* token = strtok(a_str, delim);

        while (token)
        {
            assert(idx < count);
            *(result + idx++) = strdup(token);
            token = strtok(0, delim);
        }
        assert(idx == count - 1);
        *(result + idx) = 0;
    }

    return result;
}

int main()
{
    char months[] = "JAN,FEB,MAR,APR,MAY,JUN,JUL,AUG,SEP,OCT,NOV,DEC";
    char** tokens;

    printf("months=[%s]\n\n", months);

    tokens = str_split(months, ',');

    if (tokens)
    {
        int i;
        for (i = 0; *(tokens + i); i++)
        {
            printf("month=[%s]\n", *(tokens + i));
            free(*(tokens + i));
        }
        printf("\n");
        free(tokens);
    }

    return 0;
}

Output:

$ ./main.exe
months=[JAN,FEB,MAR,APR,MAY,JUN,JUL,AUG,SEP,OCT,NOV,DEC]

month=[JAN]
month=[FEB]
month=[MAR]
month=[APR]
month=[MAY]
month=[JUN]
month=[JUL]
month=[AUG]
month=[SEP]
month=[OCT]
month=[NOV]
month=[DEC]

Answer 5

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

/**
 *  splits str on delim and dynamically allocates an array of pointers.
 *
 *  On error -1 is returned, check errno
 *  On success size of array is returned, which may be 0 on an empty string
 *  or 1 if no delim was found.  
 *
 *  You could rewrite this to return the char ** array instead and upon NULL
 *  know it's an allocation problem but I did the triple array here.  Note that
 *  upon the hitting two delim's in a row "foo,,bar" the array would be:
 *  { "foo", NULL, "bar" } 
 * 
 *  You need to define the semantics of a trailing delim Like "foo," is that a
 *  2 count array or an array of one?  I choose the two count with the second entry
 *  set to NULL since it's valueless.
 *  Modifies str so make a copy if this is a problem
 */
int split( char * str, char delim, char ***array, int *length ) {
  char *p;
  char **res;
  int count=0;
  int k=0;

  p = str;
  // Count occurance of delim in string
  while( (p=strchr(p,delim)) != NULL ) {
    *p = 0; // Null terminate the deliminator.
    p++; // Skip past our new null
    count++;
  }

  // allocate dynamic array
  res = calloc( 1, count * sizeof(char *));
  if( !res ) return -1;

  p = str;
  for( k=0; k<count; k++ ){
    if( *p ) res[k] = p;  // Copy start of string
    p = strchr(p, 0 );    // Look for next null
    p++; // Start of next string
  }

  *array = res;
  *length = count;

  return 0;
}

char str[] = "JAN,FEB,MAR,APR,MAY,JUN,JUL,AUG,SEP,OCT,NOV,DEC,";

int main() {
  char **res;
  int k=0;
  int count =0;
  int rc;

  rc = split( str, ',', &res, &count );
  if( rc ) {
    printf("Error: %s errno: %d \n", strerror(errno), errno);
  }

  printf("count: %d\n", count );
  for( k=0; k<count; k++ ) {
    printf("str: %s\n", res[k]);
  }

  free(res );
  return 0;
}

Answer 6

Method below will do all the job (memory allocation, counting the length) for you. More information and description can be found here - Implementation of Java String.split() method to split C string

int split (const char *str, char c, char ***arr)
{
    int count = 1;
    int token_len = 1;
    int i = 0;
    char *p;
    char *t;

    p = str;
    while (*p != '\0')
    {
        if (*p == c)
            count++;
        p++;
    }

    *arr = (char**) malloc(sizeof(char*) * count);
    if (*arr == NULL)
        exit(1);

    p = str;
    while (*p != '\0')
    {
        if (*p == c)
        {
            (*arr)[i] = (char*) malloc( sizeof(char) * token_len );
            if ((*arr)[i] == NULL)
                exit(1);

            token_len = 0;
            i++;
        }
        p++;
        token_len++;
    }
    (*arr)[i] = (char*) malloc( sizeof(char) * token_len );
    if ((*arr)[i] == NULL)
        exit(1);

    i = 0;
    p = str;
    t = ((*arr)[i]);
    while (*p != '\0')
    {
        if (*p != c && *p != '\0')
        {
            *t = *p;
            t++;
        }
        else
        {
            *t = '\0';
            i++;
            t = ((*arr)[i]);
        }
        p++;
    }

    return count;
}

How to use it:

int main (int argc, char ** argv)
{
    int i;
    char *s = "Hello, this is a test module for the string splitting.";
    int c = 0;
    char **arr = NULL;

    c = split(s, ' ', &arr);

    printf("found %d tokens.\n", c);

    for (i = 0; i < c; i++)
        printf("string #%d: %s\n", i, arr[i]);

    return 0;
}