Notes on MP3

CS603
University of Alabama

Spring 2004

.

The prolog program works in two phases--first it builds an abstract syntax tree (AST), then it evaluates it. The section of the program from "exp(literal(L)) ->" to "num(9)" are the rules that you must fill in to parse the program and build an AST. The section of the program from "prog_eval(exp(E), Result) :-" to the end of the file evaluates the AST "exp(E)" and binds the result as "Result"

For the language as defined, the middle section of the program, labeled ``SEMANTIC ALGEBRA'', is not necessary. If we had extended the language to include (set var exp), then we would need it. However, I will explain it here anyway. These two rules use Prolog's database to represent the store of the program. The access rule says that if there is a rule ``store(Var, Val)'' then bind Val. For example, if there is a rules in the database store(x,2)., then access(x,Result). would result in Result = 5. The update rule is a little more complicated. Let us examine the second goal first. The assert predicate places its argument into Prolog's database. In the update rule, if update(x,3) is executed, then the rule store(x,3) is placed in the database. The first goal handles the case where the rule that will be inserted by the second goal already exists. The retractall predicate removes its argument from the database. For example, if update(x,5) is entered, then any store predicate with a first argument of x would be removed.

The first thing to do is finish the code for building the AST. You can test this separately. Here is an example execution.

$ swipl
Welcome to SWI-Prolog (Multi-threaded, Version 5.2.0)
Copyright (c) 1990-2003 University of Amsterdam.
SWI-Prolog comes with ABSOLUTELY NO WARRANTY. This is free software,
and you are welcome to redistribute it under certain conditions.
Please visit http://www.swi-prolog.org for details.

For help, use ?- help(Topic). or ?- apropos(Word).

?- ['MP3.pl'].
% MP3.pl compiled 0.01 sec, 6,356 bytes

Yes
?- num(Result,[0],[]).

Result = 0 

Yes
?- literal(Result, ['#t'],[]).

Result = bool(sc_true) 

Yes
?- prim(Result,[+],[]).

Result = + 

Yes
?- literal(Result, [0], []).

Result = num(0) 

Yes
?- exp(Result, [5], []).

Result = literal(num(5)) 

Yes
?- exp(Result, ['(', if, '#t', 5, 6, ')'], []).

Result = if(literal(bool(sc_true)), literal(num(5)), literal(num(6))) 

Yes
?-
The second thing to do is to finish the code for evaluating the AST. Here is an example execution.
?- exp(P,['(', if, '#t', 5, 6, ')'], []), prog_eval(exp(P), Result).

P = if(literal(bool(sc_true)), literal(num(5)), literal(num(6)))
Result = 5 

Yes
?-

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Notes on MP3

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2004-04-22