I'm trying to define a function f, such that



Then, I tried to evaluate f[x,y]. Since f[x,y] satisfies none of these two patterns, it is expected that the result will be just f[x,y]. Nevertheless, in fact, the result is


and I get a error message:

$IterationLimit::itlim: Iteration limit of 4096 exceeded. >>

It seems quite confusing. How could this happen?

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    $\begingroup$ You're looking for the Flat attribute. $\endgroup$ – rcollyer Nov 2 '15 at 14:33
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    $\begingroup$ Also, take a look at ?f. Another way to achieve the desired result would be f[Hold@f[a__]]:=f[a];. $\endgroup$ – Graumagier Nov 2 '15 at 14:40
  • $\begingroup$ @rcollyer,@Graumagier.Thanks for all of your help. I changed my code by replacing the second line by SetAttributes[f,Flat]. However, the problem remained. $\endgroup$ – Wen Chern Nov 2 '15 at 15:20
  • $\begingroup$ I'm not really sure Flat does what you want to do. Did you try f[Hold@f[a__]]:=f[a];? Also take care to clear all definitions/restart the kernel. $\endgroup$ – Graumagier Nov 2 '15 at 15:24

Note the following



We see that the definition g[3]=2 was stored, rather than g[a]=2. The argument of g, which is a, is evaluated before the definition is made.

The same happens in your code. f[a__] evaluates to a__ before the definition is made.


I like the following solution

f4[a_] := a;
f4[HoldPattern@f4[a__]] := f4[a];

Another solution relies on using Unevaluated in a strange and AFAIK undocumented way, like this

f2[a_] := a;
f2[Unevaluated@f2[a__]] := f2[a];
f2 // Definition

I kind of like this too, because it allows you to make the definition you want to make, without HoldPattern.

| improve this answer | |
  • $\begingroup$ I have quited the kernel before runing these two lines of codes. How could the arguments be evaluated? $\endgroup$ – Wen Chern Nov 2 '15 at 15:27
  • $\begingroup$ @WenChern it is just what Mathematica does. I suppose the case of the example of g was the main case that was focused on in making this decision. $\endgroup$ – Jacob Akkerboom Nov 2 '15 at 15:33
  • $\begingroup$ The problem arises if you evaluate f[a_]=a prior to f[f[a__]]=f[a], because first f[a_] gets "replaced" with a, and then f[f[a__]]=f[a] evaluates to f[a__]=f[a]. $\endgroup$ – Graumagier Nov 2 '15 at 15:33
  • $\begingroup$ @Graumagier yes, this is essentially Chris Degnens solution. My own solution is something that may be nice to see for more experienced users. I suppose it also explicitly reminds you of the problem, which can be nice. $\endgroup$ – Jacob Akkerboom Nov 2 '15 at 15:36
  • $\begingroup$ Sure, your solution is definitely more robust. I just tried to explain the problem to Wen Chern, but your comment was put up first ;) $\endgroup$ – Graumagier Nov 2 '15 at 15:39

You can get the expected output by defining the functions in reverse order.

f[f[a__]] := f[a]
f[a_] := a

f[x, y]

f[x, y]

| improve this answer | |
  • $\begingroup$ Thank you very much. Though I don't know how this method works, it does solve the problem. $\endgroup$ – Wen Chern Nov 2 '15 at 15:37
  • $\begingroup$ @WenChern it changes the order in which the patterns are tested. Since f[f[a]] matches both f[a_] and f[f[a__]], you need to change the order they're stored in (cf. DownValues, and the answers). This is most easily effected by changing the order they're declared in. Note, however, there is a notion of specificity, where more specific patterns are tested before more general ones. In this case, though, they're equally general. $\endgroup$ – rcollyer Nov 2 '15 at 16:14
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    $\begingroup$ @rcollyer unless I am mistaken, this is not what is going on. The pattern f[f[a__]] is more specific than f[a_], so in this sense the order should not matter. In the example f7[a_] := a; f7[{a__}] := f7[a]; the DownValues are automatically sorted for this reason. The reason for the trouble is that in the other order of evaluation, there is an unexpected evaluation of a pattern. Please see my answer. $\endgroup$ – Jacob Akkerboom Nov 2 '15 at 16:40
  • $\begingroup$ @JacobAkkerboom you're correct. I misread DownValues@f, completely missing the transformation. $\endgroup$ – rcollyer Nov 2 '15 at 16:48
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    $\begingroup$ @rcollyer to be honest this is something I could easily have missed while writing code myself. It is nice to be reminded of this danger IMO. $\endgroup$ – Jacob Akkerboom Nov 2 '15 at 17:10

Jacob gives a good exposition on different methods that work. But, to avoid any possibility of ambiguity, I would go with something very different

f[a_] := a
f[q_f] := q

which is correctly ordered

(* {HoldPattern[f[q_f]] :> q, HoldPattern[f[a_]] :> a} *)
| improve this answer | |

It looks like your double uderscore (BlankSequence) in the second definition matches things like x,y. For example:

f[a__] := 10*a;
f[3, 2, 4]
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  • $\begingroup$ Sure it does, but by Wen Chern's intuition neither f[f[a__]] nor f[a_] should match f[x,y] (and they don't actually). Only in the above combination they do. $\endgroup$ – Graumagier Nov 2 '15 at 14:50
  • $\begingroup$ @Graumagier I guess I don't really understand what the desired result is. $\endgroup$ – Mitchell Kaplan Nov 2 '15 at 14:57
  • $\begingroup$ As I understand it the expression f[x,y] should be returned unevaluated because it is not of the form f[f[a__]] (a function inside a function). $\endgroup$ – Graumagier Nov 2 '15 at 14:58

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