#include <stdio.h>
#include <string.h>
int main() {
char *slogan = "together{kaliya} [namak]";
char *slow_gun = strdup(slogan);
char *token = strsep(&slow_gun, "{");
printf ("\n slow_gun: %s\n token: %s\n", slow_gun, token);
return 0;
}
when I execute it:
$ cc -o try try_strsep.c
$ ./try
slow_gun: kaliya} [namak]
token: together
But when, I change the char *slogan to:
char *slogan = "kalia} [namak]";
and execute the same program:
$ vi try_strsep.c
$ cc -o try try_strsep.c
$ ./try
slow_gun: (null)
token: kalia} [namak]
My Question is, so when I use strsep() and input string does not have the pattern I am looking for, the return of strsep() is wrong. The only way I can validate whether strsep() could not find the pattern is to check if (slow_gun == NUll)
.
If I have char *slogan = "together{"
then strsep
would successfully return token
but returns slow_gun
to blank (not null
)
$ cc -o try try_strsep.c
$ ./try
slow_gun:
token: together
Is there a way I could avoid this IF check and rely on the function to return me the substr and if its not there, return NULL
?
No, there's no way to avoid the check slow_gun == NULL
. Here's a description of strsep
's behavior:
char *strsep(char **stringp, const char *delim);
DESCRIPTION
If*stringp
isNULL
, thestrsep()
function returnsNULL
and does nothing else. Otherwise, this function finds the first token in the string*stringp
, where tokens are delimited by symbols in the stringdelim
. This token is terminated by overwriting the delimiter with a null byte ('\0'
) and*stringp
is updated to point past the token. In case no delimiter was found, the token is taken to be the entire string*stringp
, and*stringp
is madeNULL
.RETURN VALUE
Thestrsep()
function returns a pointer to the token, that is, it returns the original value of*stringp
.
So, if no match is found strsep
returns a pointer to the original string and sets the slow_gun
input to NULL.
If the delimiter is the last character in the string, that character is overwritten by '\0' and slow_gun
is set to the following character, which happens to be the '\0' terminating the original string. This is why print statement prints an empty string.
NOTE You're using strdup
incorrectly, the caller is responsible for calling free
on the pointer returned by that function.