Fungibility

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1

I don’t know if I am right, but it seems like the fungibility mechanics are broken. I tried to compile a simple example (in english) copied from the chapter 11 in the manual

(current player #player)
(#player is #in #room)
(room #room)
(fungibility enabled)

#bowl
(name *) bowl
(item *)
(container *)
(* is #in #room)
(* is handled)

(generate 5 (marble $))
(name (marble $))
    marble
(plural name (marble $))
    marbles
((marble $) is #in #bowl)
((marble $) is handled)
(item (marble $))
(fungible (marble $) (marble $))

but all I get is “You see a marble in the bowl.”
Has anybody some guessing what is wrong?

2

Well, in the example you posted, you forgot the asterisks in the bowl definition, but I’m guessing that was just a typo since you wouldn’t have gotten that response. When I compile this code:

Oops, my code only works for version 0h (sorry about that) 
(current player #player)
(#player is #in #room)
(room #room) 
(fungibility enabled)

#bowl
(name *) bowl
(item *)
(container *)
(* is #in #room)
(* is handled)

(generate 5 (marble $))
(name (marble $))	marble
(plural name (marble $))	marbles
((marble $) is #in #bowl)
((marble $) is handled)
(item (marble $))
(fungible (marble $) (marble $))

I get this output:

Location
You are here.

You see a bowl here.

> look in bowl
In the bowl you see five marbles.

> get marble
You take the marble out of the bowl.

> look in bowl
In the bowl you see four marbles.

See the real answer from Linus below.

3

Should be:

(*(marble $) is #in #bowl)
(*(marble $) is handled)
(item *(marble $))

In particular, the first line, (*(marble $) is #in #bowl) needs the asterisk starting with language version 0i. The example in the manual needs to be updated. Thanks for reporting!

2 Likes
4

It’s my logic fault. Without the asterisk the predicate succeeds only for the first “found” item which is exactly one marble. Thanks for the clarification.