I have defined a function fun[f_[x_,y_]]. When I call it, I want to wrap everything by Unevaluated:

fun[Unevaluated[f][Unevaluated[x], Unevaluated[y]]

But the above expression looks clumsy, especially if the arguments of fun is more complicated. Is there a more elegant way to do it?

A simple example of the function fun is as follows. The last line is what I expect to get.

fun[f_[x_, y_]] := ToExpression[ToString[f] <> ToString[x] <> ToString[y]]
f = 1; x = 1; y = 1;
fun[f[x, y]]
(* -> 111 *)
fun[Unevaluated[f][Unevaluated[x], Unevaluated[y]]
(* -> fxy *)

I tried several ways by using a single Unevaluated or HoldPattern, and they do not work.

  • 1
    $\begingroup$ Have you tried giving fun the HoldAll attribute? $\endgroup$ Commented Aug 14, 2015 at 16:13
  • $\begingroup$ I just tried, and it does not work. $\endgroup$
    – renphysics
    Commented Aug 14, 2015 at 16:21
  • 1
    $\begingroup$ @renphysics Why? SetAttributes[fun, HoldAll]; fun[f_[x_, y_]] := Unevaluated[f][Unevaluated[x], Unevaluated[y]]; f = x = y = 1; fun[f[x, y]] $\endgroup$
    – Kuba
    Commented Aug 14, 2015 at 16:35

1 Answer 1


If I understand what you want:

SetAttributes[fun, HoldAll];
fun[expr_] := Identity @@ Identity @@ 
 MapAll[Unevaluated, Hold[expr], Heads -> True]

The two Identity operations fix the fact that MapAll applies Unevaluated to the entire expression as well - MapAll[g, f[x,y]] is g[g[f][g[x],g[y]] when we want g[f][g[x],g[y] - and remove the Hold we artificially put in.

  • $\begingroup$ +1 for the use of MapAll! $\endgroup$
    – march
    Commented Aug 14, 2015 at 16:28

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