Implementing `strtok` whose delimiter has more than one character - c

Question

Code snippet:
char str[] = "String1::String2:String3:String4::String5";
char *deli = "::";
char *token = strtok(str,deli);
while(token != NULL)
{
printf("Token= \"%s\"\n", token);
token=strtok(NULL,deli);
}
The above code snippet produces the output:
Token="String1"
Token="String2"
Token="String3"
Token="String4"
Token="String5"
but I want the output to be:
Token="String1"
Token="String2:String3:String4"
Token="String5"
I know that I am not getting the expected output because each character in the second argument of strtok is considered as a delimiter.
To get the expected output, I've written a program that uses strstr(and other things) to split the given string into tokens such that I get the expected output. Here is the program:
#include <stdbool.h>
#include <stdlib.h>
#include <string.h>
#include <stdio.h>
int myStrtok(char* str,char* deli)
{
if(str==NULL || deli==NULL)
return -1;
int tokens=0;
char *token;
char *output=str;
while((token=strstr(output,deli))!=NULL)
{
bool print=true;
if(output != token)
{
printf("Token = \"");
tokens++;
print=false;
}
while(output != token)
{
putchar(*output);
output++;
}
if(print==false)
printf("\"\n");
output+=strlen(deli);
}
if(strlen(output)>0)
{
printf("Token = \"%s\"",output);
tokens++;
}
printf("\n\n");
return tokens;
}
int main(void)
{
char str[]="One:1:Two::Three::::";
char *deli="::";
int retval;
printf("Original string=\"%s\"\n\n",str);
if((retval=myStrtok(str,deli))==-1)
printf("The string or the delimeter is NULL\n");
else
printf("Number of tokens=%d\n", retval);
return(EXIT_SUCCESS);
}
The above program produces the expected output.
I'm wondering if there are any easier/simpler ways to do it. Are there any?

Answer 1

A string-delimiter function that uses strtok's prototype and mimicks its usage:
char *strtokm(char *str, const char *delim)
{
static char *tok;
static char *next;
char *m;
if (delim == NULL) return NULL;
tok = (str) ? str : next;
if (tok == NULL) return NULL;
m = strstr(tok, delim);
if (m) {
next = m + strlen(delim);
*m = '\0';
} else {
next = NULL;
}
return tok;
}

Answer 2

If you don't care about the same usage as strtok I would go with this:
// "String1::String2:String3:String4::String5" with delimiter "::" will produce
// "String1\0\0String2:String3:String4\0\0String5"
// And words should contain a pointer to the first S, the second S and the last S.
char **strToWordArray(char *str, const char *delimiter)
{
char **words;
int nwords = countWords(str, delimiter); //I let you decide how you want to do this
words = malloc(sizeof(*words) * (nwords + 1));
int w = 0;
int len = strlen(delimiter);
words[w++] = str;
while (*str != NULL)
{
if (strncmp(str, delimiter, len) == 0)
{
for (int i = 0; i < len; i++)
{
*(str++) = 0;
}
if (*str != 0)
words[w++] = str;
else
str--; //Anticipate wrong str++ down;
}
str++;
}
words[w] = NULL;
return words;
}

Answer 3

code derived from strsep https://code.woboq.org/userspace/glibc/string/strsep.c.html
char *strsepm( char **stringp, const char *delim ) {
char *begin, *end;
begin = *stringp;
if ( begin == NULL ) return NULL;
/* Find the end of the token. */
end = strstr( begin , delim );
if ( end != NULL ) {
/* Terminate the token and set *STRINGP past NUL character. */
*end = '\0';
end += strlen( delim );
*stringp = end;
} else {
/* No more delimiters; this is the last token. */
*stringp = NULL;
}
return begin;
}
int main( int argc , char *argv [] ) {
char *token_ptr;
char *token;
const char *delimiter = "&&";
char buffer [ 256 ];
strcpy( buffer , " && Hello && Bernd && waht's && going && on &&");
token_ptr = buffer;
while ( ( token = strsepm( &token_ptr , delimiter ) ) != NULL ) {
printf( "\'%s\'\n" , token );
}
}
Result:
' '
' Hello '
' Bernd '
' waht's '
' going '
' on '
''