I am just starting to use Prolog, and already I've run into problem with a seemingly simple example. Here is my .pl file:
hacker(P) :- mountaindew(P), doesntsleep(P).
hacker(P) :- writesgoodcode(P).
writesgoodcode(jeff).
Then, after I load the program into swipl, I test it with this line at the prompt
writesgoodcode(jeff).
I thought it would display true, but I get this error:
?- hacker(jeff).
ERROR: hacker/1: Undefined procedure: mountaindew/1
Exception: (7) hacker(jeff) ?
This program works fine, but this doesn't solve my problems:
hacker(P) :- writesgoodcode(P).
writesgoodcode(jeff).
$ swipl -s dumb.pl
% dumb.pl compiled 0.00 sec, 1,112 bytes
?- hacker(jeff).
true.
Can anyone explain why my original program doesn't work? From my understanding, Prolog should "skip" the first statement since it doesn't have enough information, and check the next line. It does have enough info for that second line, and thus it should evaluate true. Any help or a point in the right direction would be great. Thanks.
As the error message says, you have an undefined procedure mountaindew/1. To make your code return true, your options are:
Define this predicate
Declare that this predicate is dynamic: dynamic(mountaindew/1)
Declare that all unknown predicates should fail (not recommended): set_prolog_flag(unknown, fail)
you could also change the order of the predicates (cannot be done always ofc)
but mostly what Kaarel said.
in the end there is not really a point in writing something that will always fail, even if you are still developing the code
This works but as I am a beginner I can't say why. The word "uninstantiated" may apply. Despite not knowing why it works, I think it's helpful to show one way that works.
hacker(P) :- mountaindew(P), doesntsleep(P).
hacker(P) :- writesgoodcode(P).
mountaindew(john).
doesntsleep(john).
writesgoodcode(jeff).
Related
So I run into some troubles while (ab?)using
lambda.pl.
I do a "use_module(library(lambda))."
in the first lines of a file that
I consult via ["a.prolog"].
Then I get an "undefined procedure ()/3"
and some gibberish afterwards.
The same happens for any order of use_modules.
It happens whether I load a.prolog via
[...], consult or as a script from the cmdline.
I reduced the script to the currying-example from Rosseta code
https://rosettacode.org/wiki/Currying#Prolog
use_module(library(lambda)).
:- initialization(main, main).
main :-
N = 5, F = \X^Y^(Y is X+N), maplist(F, [1,2,3], L),
print(L).
It doesn't work.
It works, however, if I a manually load 'lambda'
at the swipl-prompt and immeditately consult
a.prolog. Then the goal N=5,.... works just fine.
If I, however, first consult a.prolog; then manually
use_module and then run the query, I get the error.
Reconsulting doesn't help onwards.
Somehow, the first command at the prompt needs to
be use_module.
Or do I get the loading mechanism completely wrong?
If so, please apologize; but I would love get a
hint how to solve this.
This is a common error when first using modules.
Please have a look at this line:
use_module(library(lambda)).
This is just a fact, saying "use_module(library(lambbda)) holds".
What you want instead is a directive.
A directive is a term with primary functor (:-)/1. That is, you want:
:- use_module(library(lambda)).
EDIT: For the particular case of library(lambda), I would like to add that there is a page with a lot of useful information about it that is a bit hard to find:
http://www.complang.tuwien.ac.at/ulrich/Prolog-inedit/ISO-Hiord
"Hiord" stands for higher order.
NB: Just to be clear, my motivation for the question below is to learn my way around Prolog and SWI-Prolog, not to get past a particular error message. In fact, I already know one way to get past this error. My question asks about whether several other alternatives are also possible.
An exercise in my Prolog textbook asks one to describe the outcome one should expect from several queries, assuming one has consulted the following knowledgebase beforehand:
x(a).
z(b).
z(c).
z(d).
w(V) :- x(V).
w(V) :- y(V).
w(V) :- z(V).
On SWI-Prolog, at least, most of these queries fail, because SWI-Prolog intreprets y as undefined.
From the solutions to the exercises at the end of the book I can tell that this is not the authors' intended outcome. Maybe there's a Prolog implementation for which the exercise would turn as the solution presents it.
Be that as it may, I'd like to learn about good ways to work around the problem.
Consider, for example, the query w(x).. the book's solution claims that the query w(x). should evaluate to false.. In fact, this is what happens:
?- w(x).
ERROR: w/1: Undefined procedure: y/1
Exception: (7) y(x) ?
(At this point, SWI-Prolog is expecting me to enter some letter indicating how to respond to the exception. More about this later.)
I'm looking for ways to either turn the interaction above to
?- w(x).
false.
?-
...or at least for a suitable <ONE-LETTER RESPONSE> I could give to SWI-Prolog so that it arrives at the conclusion false. IOW, so that
?- w(x).
ERROR: w/1: Undefined procedure: y/1
Exception: (7) y(x) ? <ONE-LETTER RESPONSE>
false.
?-
I know of at least one answer to my question, namely simply to delete or comment out the line:
w(V) :- y(V).
I would like to know of other possible solutions, such as, for example, the "suitable " I alluded to earlier.
Another possibility would be some SWI-Prolog global configuration that would result in the above interaction to change to
?- w(x).
false.
?-
A third possibility would be to "define" y in some minimal way. The only way I can come up with is by adding the fact
y(dummy).
to the knowledgebase. Is there a more minimal way to define y, one that does not require introducing an extraneous constant into the domain of discourse?
(This is not specific to SWI)
The first Prolog systems back in the 1970s actually behaved in the way you describe. Soon it became apparent that this is a frequent source of errors. Simple misspellings remained undetected for too long. Current implementations produce a clean existence error. This is standard behaviour since 1995.
However, you can go back into the olden tymes with the ISO Prolog flag unknown which has three values error (default), fail, and warning.
?- inex.
ERROR: Undefined procedure: inex/0 (DWIM could not correct goal)
?- set_prolog_flag(unknown, fail).
Warning: Using a non-error value for unknown in the global module
Warning: causes most of the development environment to stop working.
Warning: Please use :- dynamic or limit usage of unknown to a module.
Warning: See http://www.swi-prolog.org/howto/database.html
true.
?- inex.
false.
?- set_prolog_flag(unknown, warning).
Warning: Using a non-error value for unknown in the global module
Warning: causes most of the development environment to stop working.
Warning: Please use :- dynamic or limit usage of unknown to a module.
Warning: See http://www.swi-prolog.org/howto/database.html
true.
?- inex.
Warning: toplevel: Undefined procedure: inex/0 (DWIM could not correct goal)
false.
As you can read above, SWI proposes to use a dynamic declaration in stead - which in turn has its own problems... It is much better to declare instead:
:- discontiguous(y/1).
An undefined procedure error raises an exception so if you want the exception to be raised because you don't want to change y/1 predicate (delete or define it) you need to catch the exception and then return false like this:
x(a).
z(b).
z(c).
z(d).
w(V) :- x(V).
w(V) :- catch(y(V), error(Err,_Context),my_handler(Err)).
w(V) :- z(V).
my_handler(Err):- write(Err),fail.
Example:
?- w(x).
existence_error(procedure,y/1)
false.
I'm trying to write hasRepetition that 'recives' a list and returns true if and only if it has repetitions in it.
I wrote : hasRepetition([Head|Tail]) :- member(Head,Tail);hasRepetition(Tail).
7 ?- hasRepetition([1,1]).
ERROR: toplevel: Undefined procedure: hasRepetition/1 (DWIM could not correct goal)
Your procedure definition is fine.
I guess you forgot to consult your code, i.e. consult(name). where name is the name of the file that has your code.
I think #gusbro already give you the answer (+1). But please note that the procedure you wrote its' very inefficient.
First, memberchk/2 instead of member should be used. Then, put a cut before the disjunction. Otherwise, your code will be very slow, if called on backtracking (for instance, inside a findall...)
I wrote the following simple code, and I expect that when I write 'male.', this code ask me once "is it male?" and if i input 'No' it write on screen "she is female".
male :- ( print('is it male ? '),read(yes)) -> true; asserta( not(male)),female.
female:- not(male),print('she is female').
not(P) :- (call(P) -> fail ; true) .
but this code has following error:
uncaught exception: error(permission_error(modify,static_procedure,not/1),asserta/1);
the error in swi-prolog is :
ERROR: asserta/1: No permission to modify static_procedure `not/1'
As gusbro said, not/1 is a predefined static procedure (and therefore it is not a good idea to use the same name). However, this is not the reason you get the error in swi-prolog as you can overwrite the standard definition:
?- assert(not(42)).
true.
?- not(42).
true.
the problem is that you have already defined not/1 in your code and, when you do not declare a predicate explicitly as dynamic, swi-prolog will assume that it's static and will not allow you to change it.
You can declare it as dynamic by inserting this line in your code:
:-dynamic(not/1).
I think that this will not solve the problem in other prolog implementations (eg gnu-prolog) as the error message says permission_error(modify,static_procedure,not/1); in any case it is highly recommended to change the name.
By the way, it would be simpler and cleaner to simply test what the argument is and print the corresponding message. If, however, you want to create something more complex (using a state maybe) you could use global variables since they are more efficient that assert/retract.
I am reading http://cs.union.edu/~striegnk/learn-prolog-now/html/node3.html#subsec.l1.kb1,
but I am having trouble running the following predicate:
SICStus 4.0.1 (x86-win32-nt-4): Tue May 15 21:17:49 WEST 2007
| ?- woman(mia).
! Existence error in user:woman/1
! procedure user:woman/1 does not exist
! goal: user:woman(mia)
| ?-
If, on the other hand, I write it to a file and run consult the file, it seems to work fine...
Am I only allowed to define predicates in a file having later to consult them? Can't I just do it in the editor itself?
It's a little annoying to make predicates in the repl. You could do
| ?- ['user'].
woman(mia).
^D
ie consult user input, or
| ?- assertz(woman(mia)).
assert it. Both awkward IMO -- there might be a better way, though, I just don't know it. In general it is easier to use a script.
You should enter woman(mia). into a file to assert it as a fact. If you write it into the interpreter, it's taken as a query, not a fact.
From the SWI Prolog FAQ:
Terms that you enter at the toplevel are processes as queries, while
terms that appear in a file that is loaded into Prolog is processed as
a set of rules and facts. If a text reads as below, this is a rule.
carnivore(X) :- animal(X), eats_meat(X).
Trying to enter this at the toplevel results in the error below. Why?
Because a rule is a term :-(Head, Body), and because the toplevel
interprets terms as queries. There is no predicate with the name :-
and two arguments.
?- carnivore(X) :- animal(X), eats_meat(X). ERROR: Undefined
procedure: (:-)/2 ERROR: Rules must be loaded from a file ERROR:
See FAQ at http://www.swi-prolog.org/FAQ/ToplevelMode.txt
Isn't this stupid? Well, no. Suppose we have a term
eats_meat(rataplan). If this appears in a file, it states the fact
that rataplan eats meat. If it appears at the toplevel, it asks Prolog
to try proving whether rataplan eats meat.
If a text reads
:- use_module(library(clpfd)).
This is a directive. Directives are similar to queries, but instead of
asking the toplevel to do something, they ask the compiler to do
something. Like rules and facts, such terms belong in files.
Instead of writing to a file you can also use assert in the toplevel (as explained later in the FAQ as well).