Foldnum for integer intervals like foldl for lists

Nice to know
,
1

foldnum is a macro included in my pack package pac about ten years ago. I have fixed a bug of foldnum macro expansion related to global variable.

foldnum is a version of foldl for integer interval in stead of lists. I tried to find in SWIPL builtin library, but failed. Fortunately it is an easy routine for pac library to define such macros as the source is listed below. foldnum is handy so far, but it must have a bug for general use. If foldnum is useful and new for Prolog, I hope SWI-Prolog officially includes complete vesion as the foldl macro for lists.

Sample Use of foldnum. 
% ?- foldnum(plus, 1-10, 0, X).
%@ X = 55.

% ?- foldnum(pred([X,Y,Z]:- Z is X*Y), 1-4, 1, R).
%@ R = 24.

% ?- F = plus, foldnum(F, 1-100, 0, R).
%@ F = plus,
%@ R = 5050.

% ?- N=100,  functor(A, #, N),
%	forall(between(1, N, I), nb_setarg(I, A, I)),
%	foldnum(pred(A, ([J, C, D]:- arg(J, A, Aj), D is C * Aj) ),
%			1 - N, 1, S).
%@ N = 100,
%@ A = #(..),
%@ S = 93326215443944152681699238856266700490715968264381621468592963895217599993229915608941463976156518286253697920827223758251185210916864000000000000000000000000.
Definition of foldnum macro

Details is omitted here.

etc(foldnum, [F|As], _, meta:G, P, P):- var(F), !,
	complete_args(foldnum, [F|As], G).
etc(foldnum, [F, I-J|As], M, G, P, Q):-
	expand_arg(F, M, F0, P, P0),
	term_variables([I, J, F0], Vs),
	expand_core(
		rec(R, Vs,
			([A, U, U]:- A>J, !)
			   &
			([A, U, V]:- call(F, A, U, U0),
				A0 is A + 1,
				call(R, A0, U0, V))),
		 M, G0, P0, Q),
		 complete_args(G0, [I|As], G).
Expanded codes of a foldnum use. 
% ?- show(foldnum(plus, 1-10, 0, X)).
%@ pac:pac#68(1,0,_), where
%@ pac:pac#68(A,B,B):-pac:(A>10),!
%@ pac:pac#68(A,B,C):-pac:plus(A,B,D),pac:(E is A+1),pac:pac#68(E,D,C)
%@ X = _.
2

Why did nobody tell me that foldl is just a for-loop over a list and foldnum is a for-loop over a range of numbers? (And, I suppose for completeness, there should be a foldwhile.)
This would have made learning the concepts so much easier.

In other words, this is the imperative way of writing foldl(Pred, List, State0, StateFinal):

state = ...
for x in list:
    state = fn(state, x)
3

Yes. term_expansion for foldnum is not necessary unless good syntax and macro expansion for the closure in the first argument. As you pointed out the codes below is enough, which is clearly easy to write his own flexible way for the prologer. Now I regret a little bit to have posted such an obvious thing.

:- meta_predicate foldnum(3, ?, ?, ?).
foldnum(F, I-J, U, V):-!, foldnum(I, J, U, V, F).
foldnum(F, I..J, U, V):- foldnum(I, J, U, V, F).
%
foldnum(I, J, U, U, _):- I>J, !.
foldnum(I, J, U, V, F):- call(F, I, U, U0),
	K is I + 1,
	foldnum(K, J, U0, V, F).