How to substitue local variables into expression with free variables?

Question

I'm wondering how to do the following:

    expr = L + R;

    foo[L_, R_] := expr;

    foo[1, 2]


(* L + R *)

and have it substitute the function variables into the expression. I realize that I could create dummy variables and do it like

foo[l_, r_] := expr /. L->l /. R->r

but my actual expression has a lot of free variables, and so I would like to know if there is a simpler way.

foo[L_,R_]=expr; The := is delayed evaluation which is usually correct but not in this precise case. — Ymareth, Jan 21 '14 at 18:30
Use = instead of :=. It's not necessary to Evaluate IMO because it would be completely equivalent to just using =. I'd consider this question a duplicate of What is the difference between Set and SetDelayed?. — Szabolcs, Jan 21 '14 at 19:30

score 1 · Accepted Answer · edited Jan 21 '14 at 19:32

1

As noted in the comments, use Set (=) instead of SetDelayed (:=) while making sure that L and R have no value assigned:

expr = L+R
foo[L_, R_] = expr

foo[1, 2]
(* ==> 3 *)

edited Jan 21 '14 at 19:32

Szabolcs

answered Jan 21 '14 at 19:30

Ajasja

I went ahead and changed this, since it was community wiki. I think Evaluate would be completely equivalent to just using Set here. Hope you don't mind, please check the edit. – Szabolcs Jan 21 '14 at 19:32
@Szabolcs Sure, even if these were not CW I welcome good edits:) – Ajasja Jan 21 '14 at 20:13

1 Answers1