Prolog if statement without else

Question

I have seen similar previous posts, but I seem to still be stuck. My current if statement, even when it fails, outputs the "if true" code. So I'd like to ask what the proper syntax is if I don't want to do anything else if the condition turns to be false.

My code that fails:

(Disease == malaria
        ->  (patient_age(y)
            ->  writeln('Since patient is young, likelihood of malaria increases.')
            ;   true)
        ;
Disease == tuberculosis
        ->  (patient_age(e)
            ->  writeln('Since patient is old, likelihood of tuberculosis increases.')
            ;   true)
).

What happens is that, even if patient_age() is false, the writeln() statement is still executed. I have attempted replacing the "false code" with another writeln() statement, which did not display. I apologize that this has been asked before. Thank you very much.

Paulo Moura Paulo Moura · Accepted Answer · 2021-02-10T13:47:40

It would be more declarative, idiomatic, and clear to rewrite the code to avoid the use of the if-then-else control construct. For example:

% likelihood(Disease, PatientAge)
likelihood(malaria, y) :-
    writeln('Since patient is young, likelihood of malaria increases.').
likelihood(tuberculosis, e) :-
   writeln('Since patient is old, likelihood of tuberculosis increases.').

Still, it would likely be best to move the written messages to a predicate that calls the likelihood/2 predicate. For example:

print_diagnostic(PatientAge) :-
    likelihood(Disease, PatientAge),
    write('Increased likelihood of '), write(Disease), nl.

Prolog if statement without else

1 Answers