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Succeeds if both terms are bound.  The value of the term is X + Y.
Use with C<is(X,Y)>.

 is(X, plus(N,1)).

=item print/1

Prints the current Term.  If the term is an unbound variable, it will print the
an underscore followed by the internal variable number (e.g., "_284").

 print(ovid).         % prints "ovid"
 print("Something").  % prints "Something"
 print(Something).    % prints whatever variable Something is bound to 

=item println/1

Same as C<print(Term)>, but automatically prints a newline at the end.

=item pow/2

Succeeds if both terms are bound.  The value of the term is X ** Y 
(X raised to the Y power).
Use with C<is(X,Y)>.

=item retract/1

Remove facts from the database.  You cannot remove rules.  This may change in
the future.  See C<assert(X)>.

 retract(loves(ovid,java)).

=item trace/0

Turns on tracing of Prolog's attempt to satisfy goals.

=item true/0

True goal.  Automatically succeeds.

=item var/1

Succeeds if X is an unbound variable.  Otherwise, this goal fails.

=item write/1

Prints the current Term.  If the term is an unbound variable, it will print the
an underscore followed by the internal variable number (e.g., "_284").

 write(ovid).         % prints "ovid"
 write("Something").  % prints "Something"
 write(Something).    % prints whatever variable Something is bound to 

=item writeln/1

Same as C<write(Term)>, but automatically prints a newline at the end.

=back

=head1 LIMITATIONS

These are known limitations that I am not terribly inclined to fix.  See the
TODO list for those I am inclined to fix.

 IF -> THEN; ELSE not allowed.

Use C<if(IF, THEN, ELSE)> instead.

Chaining terms with a semicolon for "or" does not work.  Use C<or/2> instead.

=head1 TODO

There are many things on this list.  The core functionality is there, but I do
want you to be aware of what's coming.

=over 4

=item Improve printing.

There are some bugs with printing and escaping characters.  Maybe I'll look
into them :)

=item More builtins.

Currently, we only have a tiny subset of builtins available.  More are coming.

=back

=head1 MATH

Since version .70, math is fully available in C<AI::Prolog>.  Note that math is
implemented via the
L<AI::Prolog::Parser::PreProcessor::Math|AI::Prolog::Parser::PreProcessor::Math>
module.  This module rewrites Prolog math to an internal, predicate-based form
with the L<AI::Prolog::Parser|AI::Prolog::Parser> can parse.  This may cause
confusion when debugging.

 X is 5 + 7.
 % internally converted to 
 % is(X, plus(5, 7)).

The math predicates are officially deprecated and I<cannot> be used in the same
expression with regular Prolog math.  

Number may be integers, floats, doubles, etc.  A number that starts with a
minus sign (-) is considered negative.  No number may end in a decimal point as
the period is interpreted as the end of a clause.  The following is therefore a
syntax error:

 X is 5. + 7.

Unfortunately, the parser doesn't yet yell about that.  We'll try and figure
out why later.

Omit the period after the number or put a zero after it:

 X is 5.0 + 7.
 X is 5 + 7.

Because numbers use Perl scalars, you may mix types (ints and floats) and they
will behave as you expect in Perl.

Precedence is C<*> and C</>, left to right, followed by C<+> and C<->, left to
right followed by C<%>, left to right.  (I probably should change that.)
Naturally, parentheses may be used for grouping:

 X is 3 * 5 + 2.   % is(X, plus(mult(3, 5), 2)).
 X is 3 * (5 + 2). % is(X, mult(3, plus(5, 2))).

When using math, note that C<is> is similar to Perl's assignment operator, C<=>.
This can be confusing.

 X is 3 + 2.

Sets C<X> to the value of C<5>.  

If C<X> is already instantiated, this goal succeeds if the value of C<X> is the
value of the result of the right-hand side of the equation.  Internally, if X is not
instantiated, it looks like this:

 is(5, plus(3,2)).

The C<=> operator tries to unify the left-hand side with the right-hand side:

 X = 3 + 2.

If C<X> is already instantiated, this goal succeeds if the value of C<X> is the
same goal as the right-hand side of the equation.  Internally, if X is not
instantiated, it looks like this:

 eq(plus(3,2), plus(3,2)).

When you first start using Prolog, you probably was C<is> instead of C<=>.

Logical comparisons are straightforward:

 3 >= X.
 Y > (4 + 3) * X.
 X == Y. % a test for equality
 X \= Y. % Not equal.  See caveats for ne/2
 % etc.

=head1 BUGS

None known.

=head1 SEE ALSO