Question

On my system, the following program:

int main(){
        char *strgptr;
        char buf[5] = {'b','a','a','a','\0'};
        char *tmp = strtok_r(buf, ".", &strgptr);
        if(tmp != NULL){
                printf("Found a . in baaa?\n");
                printf("It was found starting at: %s\n", tmp);
        }
        else
                printf("Everything is working.\n");
}

prints:

Found a . in baaa?
It was found starting at: baaa

However, if I swap the “.” delimiter string in strtok_r for “a”, I get (as expected):

Found a . in baaa?
It was found starting at: b

But swapping the “.” for any other char not appearing in buf (e.g. “c”) produces:

Found a . in baaa?
It was found starting at: baaa

The man page for strtok_r, as expected, says:

The strtok() and strtok_r() functions return a pointer to the next token, 
or NULL if there are no more tokens.

So why does strtok_r fail to return NULL when passed a string which contains none of the tokens in question?

Answer 1

Because the delimiter isn’t found, so you’re getting the whole string returned. It acts as if there’s an invisible delimiter after the string.