function split which I want to include the delimiter " : " - c

Question

I have this function which returns an array of strings in C which are delimited by delim. Every call to strtok returns the string between two delimiters, but it doesn't include the delimiter. I want it to return the string with the char of delim at the end only if this char==':'. For example, for " ici: papa mama", if I call it with delim== " \t\n:" I want it to return an array which contains:
ici:
papa
mama
(the ':' will be included in ici but not the spaces or tabulations).
need help
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
char ** split(const char * str, const char * delim)
{
/* count words */
char * s = strdup(str);
char* inter=malloc(sizeof(char)*100);// pour l etiquette ne pas oublier de mettre deux points a la fin avant de la mettre dans le
if (strtok(s, delim) == 0){
/* no word */
return NULL;}
int nw = 1;
while (strtok(NULL, delim) != 0)
nw += 1;
strcpy(s, str); /* restore initial string modified by strtok */
/* split */
char ** v = malloc((nw + 1) * sizeof(char *));
int i;
v[0] = strdup(strtok(s, delim));
for (i = 1; i != nw; ++i)
v[i] = strdup(strtok(NULL, delim));
v[i] = NULL; /* end mark */
free(s);
return v;
}
int main() {
char **words=split("ici: haha papa", " \t\n:");
printf("%s \n",words[0]);// with the two points
printf("%s \n",words[1]);// without the spaces
printf("%s \n",words[2]);// without the spaces
return 0;

Answer 1

For a generalized solution, the only way to preserve a single separator/delimiter would be to write your own tokenizer.
In your specific case, however, it would appear that the colon ‘:’ only appears attached to the end of the first token. strtok() makes this kind of issue particularly easy, since you can change the separator/delimiter list with each invocation.
char * first = strtok( s, " \t\n\r" ); // first token ends with whitespace
For the remaining tokens, get them in a loop with the colon included in the list of separators/delimiters.
char * tok = strtok( NULL, ": \t\n\r" );
As per your added comment:
For the case where there may be more than one space-delimited word before the first colon (or, as per your comment, where the colon does not have adjacent whitespace), you will have to use strchr() (or some other search function) to find the token and feed the two halves of the string into separate tokenization loops.
Error handling
Remember to look closely at the documentation when designing your code. C makes it look like you can chain stuff together nicely, but many times you cannot (without extra helper functions). For example, strtok() may return NULL, which is not a valid input to strdup() (causing UB).
In other words, make sure to check for error conditions with the return values from function invocations.
Replacing strtok
And back to the first idea, strtok() is likely implemented in terms of the strspn() and strcspn() functions. Rather than use strtok(), use those, and use strndup() to get desired substrings.
UPDATE
I was bored, so... Here is a generalized tokenizer that might do something like what you want...
#include <iso646.h>
#include <stdlib.h>
#include <string.h>
char * tok2( const char ** s, const char * ks, const char * ds )
{
// s may not be NULL
// *s may be NULL
if (!*s) return NULL;
// skip any leading ds
const char * first = *s += strspn( *s, ds );
if (!*first) return NULL;
// find the first ks|ds
while (**s and !strchr( ks, **s ) and !strchr( ds, **s )) *s += 1;
// skip any ks
*s += strspn( *s, ks );
// return substring from first (inclusive) to *s (exclusive)
size_t n = *s - first;
if (!n) return NULL;
char * result = malloc( n+1 );
if (!result) return NULL;
result[n] = '\0';
return strncpy( result, first, n );
}
It compiles cleanly with MSVC and Clang. On Windows don’t forget to add -D_CRT_SECURE_NO_WARNINGS to your command-line options to silence the stupid.
Here is the very simple test I used for it (just concatenate this to the above and compile):
#include <stdio.h>
int main(void)
{
const char * test_string = " one two: three four:five six : seven eight :nine ten ";
printf( "given: \"%s\"\n", test_string );
const char ** src = &test_string;
char * str;
while ((str = tok2( src, ":", " \t\r\n" )))
{
printf( " \"%s\"\n", str );
free( str );
}
}