Implementing `strtok` whose delimiter has more than one character

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;
}