pl/sql - strip de caracteres en password

[danyesu]

Que raro... este es el script que estoy corriendo:

Código:

- This script sets the default password resource parameters
-- This script needs to be run to enable the password features.
-- However the default resource parameters can be changed based 
-- on the need.
-- A default password complexity function is also provided.
-- This function makes the minimum complexity checks like
-- the minimum length of the password, password not same as the
-- username, etc. The user may enhance this function according to
-- the need.
-- This function must be created in SYS schema.
-- connect sys/<password> as sysdba before running the script

CREATE OR REPLACE FUNCTION verify_function
(username varchar2,
  password varchar2,
  old_password varchar2)
  RETURN boolean IS 
   n boolean;
   m integer;
   differ integer;
   isdigit boolean;
   ischar  boolean;
   ispunct boolean;
   digitarray varchar2(20);
   punctarray varchar2(25);
   chararray varchar2(52);

BEGIN 
   digitarray:= '0123456789';
   chararray:= 'abcdefghijklmnopqrstuvwxyzABCDEFGHIJKLMNOPQRSTUVWXYZ';
   punctarray:='.!"#$%&()``*+,-/:;<=>?_';

   -- Check if the password is same as the username
   IF NLS_LOWER(password) = NLS_LOWER(username) THEN
     raise_application_error(-20001, 'Password same as or similar to user');
   END IF;

   -- Check for the minimum length of the password
   IF length(password) < 4 THEN
      raise_application_error(-20002, 'Password length less than 4');
   END IF;
   
   -- Check for the maximum length of the password
   IF length(password) > 8 THEN
	  	password := SUBSTR(password,1,8);
      --raise_application_error(-20002, 'Password length more than 8');
   END IF;

   -- Check if the password is too simple. A dictionary of words may be
   -- maintained and a check may be made so as not to allow the words
   -- that are too simple for the password.
   IF NLS_LOWER(password) IN ('welcome', 'database', 'account', 'user', 'password', 'oracle', 'computer', 'abcd') THEN
      raise_application_error(-20002, 'Password too simple');
   END IF;

  
   -- Check if the password differs from the previous password by at least
   -- 3 letters
   IF old_password IS NOT NULL THEN
     differ := length(old_password) - length(password);

     IF abs(differ) < 3 THEN
       IF length(password) < length(old_password) THEN
         m := length(password);
       ELSE
         m := length(old_password);
       END IF;

       differ := abs(differ);
       FOR i IN 1..m LOOP
         IF substr(password,i,1) != substr(old_password,i,1) THEN
           differ := differ + 1;
         END IF;
       END LOOP;

       IF differ < 3 THEN
         raise_application_error(-20004, 'Password should differ by at \
         least 3 characters');
       END IF;
     END IF;
   END IF;
   -- Everything is fine; return TRUE ;   
   RETURN(TRUE);
END;

Con ese codigo agregado password := SUBSTR(password,1,8); me tira error de compilacion, pero sin ese codigo funciona bien (pero sin cumplir la condicion que necesito claro..)