!  Identical decides whether or not two objects can be distinguished from
!  each other by anything the player can type.  If not, it returns true.
! ----------------------------------------------------------------------------

[ Identical o1 o2 p1 p2 n1 n2 i j flag;
    if (o1 == o2) rtrue;  ! This should never happen, but to be on the safe side
    if (o1 == 0 || o2 == 0) rfalse;  ! Similarly
    if (parent(o1) == compass || parent(o2) == compass) rfalse; ! Saves time

    !  What complicates things is that o1 or o2 might have a parsing routine,
    !  so the parser can't know from here whether they are or aren't the same.
    !  If they have different parsing routines, we simply assume they're
    !  different.  If they have the same routine (which they probably got from
    !  a class definition) then the decision process is as follows:
    !
    !     the routine is called (with self being o1, not that it matters)
    !       with noun and second being set to o1 and o2, and action being set
    !       to the fake action TheSame.  If it returns -1, they are found
    !       identical; if -2, different; and if >=0, then the usual method
    !       is used instead.

    if (o1.parse_name ~= 0 || o2.parse_name ~= 0) {
      if (o1.parse_name ~= o2.parse_name) rfalse;
      parser_action = ##TheSame; parser_one = o1; parser_two = o2;
      j = wn; i = RunRoutines(o1,parse_name); wn = j;
      if (i == -1) rtrue;
      if (i == -2) rfalse;
    }

    !  This is the default algorithm: do they have the same words in their
    !  "name" (i.e. property no. 1) properties.  (Note that the following allows
    !  for repeated words and words in different orders.)

    p1 = o1.&1; n1 = (o1.#1)/WORDSIZE;
    p2 = o2.&1; n2 = (o2.#1)/WORDSIZE;

    !  for (i=0 : i<n1 : i++) { print (address) p1-->i, " "; } new_line;
    !  for (i=0 : i<n2 : i++) { print (address) p2-->i, " "; } new_line;

    for (i=0 : i<n1 : i++) {
        flag = 0;
        for (j=0 : j<n2 : j++)
            if (p1-->i == p2-->j) flag = 1;
        if (flag == 0) rfalse;
    }

    for (j=0 : j<n2 : j++) {
        flag = 0;
        for (i=0 : i<n1 : i++)
            if (p1-->i == p2-->j) flag = 1;
        if (flag == 0) rfalse;
    }

    !  print "Which are identical!^";
    rtrue;
];