How do I print dynamic/1 in Prolog - prolog

Question

I want to print these all names under each dynamic/1. For example I want to print all names of male under one query for print all males in Prolog.

Answer 1

The easiest way to do this in swi-prolog (and also in gnu-prolog) is to use predicate listing/1:
?- listing(male/1).
:- dynamic male/1.
male(saad).
male(sohaib).
male(salman).
...
true.
If you want more control over display formatting, you can define your own predicate as:
list_facts(Name/Arity) :-
functor(Head, Name, Arity),
forall( clause(Head, true),
format('~w.\n', [Head]) ). % Adjust formatting here!
Example:
?- list_facts(female/1).
female(rida).
female(florida).
female(yasmeen).
...
true.
If you want to show just the names, try:
?- forall(female(Name), format('~w\n', [Name])).
rida
florida
yasmeen
...
true.
REMARK The swi-prolog built-in predicate forall/2is defined as:
forall(Cond, Action) :-
\+ (Cond, \+ Action).

Answer 2

The supplied code works with SWI-Prolog, don't know if this works with other Prolog systems but the basis of the code should be portable.
This is from actual code I use daily so it more production quality code than simple answer code but the basis is to use functor/3 to get a head and then use the head in forall/2 to get the rules (rule/2) and format/3 to output each rule. Normally a rule is a head and body but since this is outputting facts only the heads are output. To convert the rules to individual values =../2 is used. This uses library(options). This also uses current_predicate/1 to check if the predicate exists to avoid an exception if it does not. functor/3 could probably be used but somewhere I recall a corner case that it did now work. The rest should not need explaining.
I only entered 3 of the male facts. Would have used all of them if they were typed as text into the question as then they could have been easily copied.
If the facts are in a different module remember to change the name of the module when calling list_facts/N.
:- module(examples,
[
list_facts/1,
list_facts/2,
list_facts/3
]).
:- dynamic female/1.
:- dynamic male/1.
male(saad).
male(sohaib).
male(salman).
list_facts(Module:Name/Arity) :-
list_facts(user_output,Module:Name/Arity,[]).
list_facts(Output_stream,Module:Name/Arity) :-
list_facts(Output_stream,Module:Name/Arity,[]).
list_facts(Output_stream,Module:Name/Arity,Options) :-
(
current_predicate(Module:Name/Arity)
->
(
option(left_margin_size(Left_margin_size),Options)
->
atomic_list_concat(['~|~',Left_margin_size,'+~|~w'],Format)
;
Format = '~w'
),
(
option(title(Title),Options)
->
format(Output_stream,Format,[Title]),
format(Output_stream,'~n',[])
;
true
),
functor(Head,Name,Arity),
forall(
rule(Module:Head,Rule),
list_fact(Output_stream,Format,Rule)
)
;
format(Output_stream,'No facts found: ~w:~w/~w.~n',[Module,Name,Arity])
).
list_fact(Output_stream,Format,Fact) :-
Fact =.. Fact_values,
list_values(Output_stream,Format,false,Fact_values).
% Recursive case - functor name
list_values(Output_stream,Format,false,[_Value|Values]) :-
!,
list_values(Output_stream,Format,true,Values).
% Recursive case - fact arguments
list_values(Output_stream,Format,true,[Value|Values]) :-
!,
format(Output_stream,Format,[Value]),
list_values(Output_stream,Format,true,Values).
% Base case
list_values(Output_stream,_Format,_Functor_done,[]) :-
format(Output_stream,'~n',[]).
Example usage.
?- working_directory(_,'C:/Users/Groot').
true.
?- [examples].
true.
?- list_facts(examples:male/1).
saad
sohaib
salman
true.
?- list_facts(user_output,examples:male/1,[left_margin_size(3),title('male/1')]).
male/1
saad
sohaib
salman
true.