Prolog, counting from an interval; (SWI-PROLOG)

I have a small question. I need to make a predicate that counts from a natural number to some other natural number. I have to implement a check too, that the second interval is bigger than the first one. However I got stuck during my way.

Here is my code (SWI-PROLOG)

count(O, _开发者_如何学JAVA, O).
count(A, B, C) :- count(A, B, D), C is D+1, C =< B.

It works kind of well as, I can get the results C=1, C=2, C=3, C=4 if I type in count(1, 4, C). However I get stuck at the end, it will result in an error with stack overflow.

The question is how do I make it to stop? I have tried almost everything. =(

Thanks for your response!

SWI-Prolog has a builtin for that...

?- help(between).
between(+Low, +High, ?Value)
    Low  and High are  integers, High >=Low.   If  Value is an  integer,
    Low=< Value=< High.   When Value  is a  variable it is  successively
    bound  to  all integers  between  Low and  High.    If High  is  inf
    or  infinite between/3 is  true iff Value>= Low,  a feature that  is
    particularly  interesting  for generating  integers from  a  certain
    value.

true.

?- between(1, 4, Value).
Value = 1 ;
Value = 2 ;
Value = 3 ;
Value = 4.

?- 

As Paulo Moura pointed out, re-ordering will solve part of the problem. Getting it to terminate gracefully, may be achieved by adding an additional clause to handle the recursive terminating condition.

Try this.

countAtoB(A,B,A) :-
    A =:= B, !.

countAtoB(A,B,A) :-
    A < B.

countAtoB(A,B,I) :- 
    A < B, 
    X is A+1,  
    countAtoB(X,B,I).

Queries would then look like this. For the sake of comparison, I have repeated the same set of test queries using between/3, strait after.

?- countAtoB(1,4,I).
I = 1 ;
I = 2 ;
I = 3 ;
I = 4.

?- countAtoB(4,4,I).
I = 4.

?- countAtoB(4,1,I).
false.

?- countAtoB(4,1,1000).
false.

?- between(1,4,I).
I = 1 ;
I = 2 ;
I = 3 ;
I = 4.

?- between(4,4,I).
I = 4.

?- between(4,1,I).
false.

?- between(4,1,1000).
false.

?-

This

count(A, B, C) :- count(A, B, D), ...

causes an infinite recursion.

Just reorder the predicate, like this:

count(A, B, A) :- A =< B.
count(A, B, C) :- A < B, A2 is A+1, count(A2, B, C).