strtol()

[Pavel Krcmar]

Hi,
I'am a little bit confused. I tried this snippet (below) and ( endptr !=
NULL ) is true everytime. I was looking to
http://www.mkssoftware.com/docs/man3/strtol.3.asp
and then to
http://www.cplusplus.com/ref/cstdlib/strtol.html
The second www specifies things like they are done - endptr is not NULL.
Is something wrong?

Best Regards,
Pavel Krcmar

---------------------------------
int main()
{
int i;
const char *z = "345";
char *endptr;

i = strtol( z, &endptr, 10 );
if ( endptr != NULL )
{
printf("endptr is not zero!\n");
printf("%s\n", endptr );
}
printf( "i=%i\n", i );
return 0;
}
---------------------------------

[Sam Holden]

On Wed, 01 Sep 2004 10:42:25 +0200,
Pavel Krcmar <pa**********@siemens.com> wrote:

Hi,
I'am a little bit confused. I tried this snippet (below) and ( endptr !=
NULL ) is true everytime. I was looking to
http://www.mkssoftware.com/docs/man3/strtol.3.asp
and then to
http://www.cplusplus.com/ref/cstdlib/strtol.html
The second www specifies things like they are done - endptr is not NULL.
Is something wrong?
endptr simply gets set to point to the first character in the string
that is not part of the number. It will never be set to NULL, if
you pass NULL instead of an actual pointer to a char* then this
won't happen.

Best Regards,
Pavel Krcmar

---------------------------------
int main()
{
int i;
const char *z = "345";
char *endptr;

i = strtol( z, &endptr, 10 );
if ( endptr != NULL )
{
printf("endptr is not zero!\n");
printf("%s\n", endptr );
}
printf( "i=%i\n", i );
return 0;
}
---------------------------------

endptr will never equal NULL in that code, since it will be
set to point to the trailing '\0' in "345" by strtol.

--
Sam Holden