How does printf know the end of the tokens of strtok while printing them out?

Question

I'm just trying to figure out what's going on in the background, when I tokenize a c string with strtok and print out the tokens by printf with %s.

So this is the example:

char str[] = "Where - is - the - end - of - tokens?";
const char s[2] = "-";
char *token;

/* get the first token */
token = strtok(str, s);

/* walk through other tokens */
while( token != NULL ) {
    printf( " %s\n", token );
    token = strtok(NULL, s);
}

return(0);

My beliefs: - strtok doesn't create a copy of the given c string, just return the memory address of the first character of the given c string. - printf with %s will print out characters from a memory address until /0.

My question is the following: How does printf know where to stop printing out characters of a token with %s?

Please help my understand this behaviour of printf with %s.

Read the docs for strtok again, you only have half the story. — Mat, Dec 24 '18 at 08:34
It replaces every sequence of tokens with --- what? (hint: that's how it knows where the token ends) — David C. Rankin, Dec 24 '18 at 09:04
Please see this: https://stackoverflow.com/questions/21097253/how-does-the-strtok-function-in-c-work — manoliar, Dec 24 '18 at 12:18
Change `printf( " %s\n", token );` to `printf( " \n", token, str );` for a more informative output. — chux - Reinstate Monica, Dec 24 '18 at 13:48

score 1 · Accepted Answer · answered Jan 02 '19 at 18:27

1

I've got my answer what is so simple: "Each delimiter is replaced by a null character." Source: IBM documentation

answered Jan 02 '19 at 18:27

Sami

11
4

How does printf know the end of the tokens of strtok while printing them out?

1 Answers1