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Browsing parserm.h Identical (lines 2732-2787) 2732 ! Identical decides whether or not two objects can be distinguished from 2733 ! each other by anything the player can type. If not, it returns true. 2734 ! ---------------------------------------------------------------------------- 2735 2736 [ Identical o1 o2 p1 p2 n1 n2 i j flag; 2737 2738 if (o1==o2) rtrue; ! This should never happen, but to be on the safe side 2739 if (o1==0 || o2==0) rfalse; ! Similarly 2740 if (parent(o1)==compass || parent(o2)==compass) rfalse; ! Saves time 2741 2742 ! What complicates things is that o1 or o2 might have a parsing routine, 2743 ! so the parser can't know from here whether they are or aren't the same. 2744 ! If they have different parsing routines, we simply assume they're 2745 ! different. If they have the same routine (which they probably got from 2746 ! a class definition) then the decision process is as follows: 2747 ! 2748 ! the routine is called (with self being o1, not that it matters) 2749 ! with noun and second being set to o1 and o2, and action being set 2750 ! to the fake action TheSame. If it returns -1, they are found 2751 ! identical; if -2, different; and if >=0, then the usual method 2752 ! is used instead. 2753 2754 if (o1.parse_name~=0 || o2.parse_name~=0) 2755 { if (o1.parse_name ~= o2.parse_name) rfalse; 2756 parser_action=##TheSame; parser_one=o1; parser_two=o2; 2757 j=wn; i=RunRoutines(o1,parse_name); wn=j; 2758 if (i==-1) rtrue; if (i==-2) rfalse; 2759 } 2760 2761 ! This is the default algorithm: do they have the same words in their 2762 ! "name" (i.e. property no. 1) properties. (Note that the following allows 2763 ! for repeated words and words in different orders.) 2764 2765 p1 = o1.&1; n1 = (o1.#1)/2; 2766 p2 = o2.&1; n2 = (o2.#1)/2; 2767 2768 ! for (i=0:ii, " "; } new_line; 2769 ! for (i=0:ii, " "; } new_line; 2770 2771 for (i=0:i<n1:i++) 2772 { flag=0; 2773 for (j=0:j<n2:j++) 2774 if (p1-->i == p2-->j) flag=1; 2775 if (flag==0) rfalse; 2776 } 2777 2778 for (j=0:j<n2:j++) 2779 { flag=0; 2780 for (i=0:i<n1:i++) 2781 if (p1-->i == p2-->j) flag=1; 2782 if (flag==0) rfalse; 2783 } 2784 2785 ! print "Which are identical!^"; 2786 rtrue; 2787 ];

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