The built-in Prolog predicate 'clause' extracts the structure of defining clauses from memory. For example, suppose that the file lists.pro has been loaded. Then the following behavior can be observed (if 'member/2' is dynamic):
?- clause(member(A,B),Body). A = X1 B = [X1,_] Body = true ; A = X2 B = [_ | X3] Body = member(X2,X3) ; no
To check that this is the correct answer look back at the definition of 'member' in the file lists.pro. Note that the two answers correspond to the two defining clauses for 'member'. The 'X1', 'X2', and 'X3' are representations for "internal" Prolog variables (these differ from system to system).
Note that in a goal of the form 'clause(H,T)' H must be instantiated enough so that the main predicate of the clause can be determined. All other bindings could be made through general unification.
As a further example, here is a Prolog program using the 'clause' predicate that analyzes a program (given as a list of predicates) to determine which predicates require which other predicates in their definition.
:- dynamic calltree/1,analyze_all/2,analyze/2,process/3, uses/2,record/3,calls/1,display_dependencies/1,forget/0 . calltree(Program) :- analyze_all(Program,Program), display_dependencies(Program), forget. analyze_all([Pred|Rest],Program) :- analyze(Pred,Program), analyze_all(Rest,Program). analyze_all(,_). analyze(P/N,Program) :- functor(PE,P,N), /* generate predicate expression */ clause(PE,Body), /* fetch definition */ process(P/N,Body,Program), /* Pred calls Body literals */ fail. analyze(_,_). /* Force bactracking then succeed */ process(P,(Q,Rest),Program) :- record(P,Q,Program), process(P,Rest,Program). process(P,(Q),Program) :- record(P,Q,Program). record(P,Q,Program) :- functor(Q,QF,N), member(QF/N,Program), \+uses(P,QF/N),!, assertz(uses(P,QF/N)). record(_,_,_). /* already recorded */ display_dependencies([P|R]) :- calls(P), nl, display_dependencies(R). display_dependencies(). calls(P) :- write(P), write(' calls: '), uses(P,Q), write(Q), write(' '), fail. calls(_). forget :- retract(uses(_,_)), fail. forget. /* use this program on itself ... */ go :- calltree([calltree/1,analyze_all/2,analyze/2,process/3, record/3,calls/1,display_dependencies/1,forget/0,uses/2).
The dynamic declaration is required so that when the program is loaded, its clauses will be stored in a form accessible to the predicate 'clause'.
To illustrate, the last definition in the file calltree.pro sets up a self-analysis of the program.
?- go. calltree/1 calls; analyze_all/2 display_dependencies/1 forget/0. analyze_all/2 calls; analyze/2 analyze_all/2 analyze/2 calls: process/3 record/3 calls: calls/1 calls: uses/2 display_dependencies/1 calls: calls/1, display_dependencies/1 forget/0 calls: uses/2 calls: yes
Let us inspect a clause tree for with root 'analyze(calltree/1,[...])' where the list [...] is the one referred to in the program.
Note how the built-in 'functor' works:
?- functor(E,calltree,1). E = calltree(X1) ?- functor(calltree(X),P,N) P = calltree N = 1
An interesting observation is that the program clause tree above cannot be extended so as to have all true leaves, since the rightmost leaf is 'fail'. The purpose of 'fail' in the program clause is to force backtracking for the built-in 'clause' so that all clauses are processed. Actually, 'analyze(calltree/1,[...])' is a consequence of the program using the second 'analyze' clause. Thus, all of the "work" is done by forcing failure. This aspect of Prolog is called failure- driven backtracking. The problem with it involves the observation just made about the clause tree: The tree captures the meaning of the program, showing where the "work" is done, but that tree does not itself establish that the goal is a consequence of the program.
Prolog sequences are processed somewhat differently than lists. The reader should do some experimenting with sequences, such as the following:
?- (X,R) = (a,b,c) X = a R= (b,c) ?- (X) = a X = a ?- (X,R) = a no
The last goal illustrates the fact that there is no explicit "empty" sequence representation (as for lists), only a representation for a sequence with just one thing left in it. The student should read the program definition for 'process' to see how it processes the literals in a body (given as the second argument).
Exercise 2.18.1 Find a way to avoid the failure-driven backtracking. Your answer may depend on what kind of Prolog is being used.
Exercise 2.18.2 Note that the program ignores literals embedded within negation. For example, the sample response did not consider 'uses/2' as being called by 'record/3' because the occurrence of 'uses/2' in 'record/3' took the form '\+uses(...)'. Modify the program so that literals within negation are considered.
Exercise 2.18.3 Modify the calltree program so that the program clauses to be analyzed are read from a file and not asserted into memory.