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I have a recursive function that does some work to an object for a repeatedamount of times.

f[h_, repeated_Integer /; repeated >= 0, level_Integer /; level > 0] :=
 Module[{},
  (* Do some stuff to h using level then call foo on h *)
  foo[h, If[level < repeated, f[h, repeated, level + 1], final[h] ]]
]

I mostly pass in 1 for the level parameter but there are times where I want to start at a different level (aka depth) of h. I found I could use the longhand Optional to skip passing in 1 for level all the time.

Optional[level_Integer /; level > 0 , 1]

However, before this I tried using shorthand optional : with no success.

(level_Integer : 1) /; level > 0

and

(level_Integer /; level > 0) : 1

Both of these give syntax errors. Is there a way to do this with shorthand optional :?

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    $\begingroup$ I though it should have duplicate but I only found this related question about PatternTest+Optional. 4937 $\endgroup$
    – Kuba
    Commented Apr 9, 2015 at 10:14
  • $\begingroup$ @Kuba I see. I didn't realise you could condition between the function signature and the SetDelayed. Thanks. I think I'll leave this one with longhand if this is the only option. I think it is cleaner. However I now can add some parameter interaction test with this new found technique. Like g[x_Integer /; x >= 0, y_Integer /; y > 0] /; y > x := {x , y}. Now that is powerful stuff. :-) $\endgroup$
    – Edmund
    Commented Apr 9, 2015 at 10:17

1 Answer 1

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I think you can move condition outside brackets:

ClearAll[f]; 
f[x_Integer : 1] /; x >= 0 := {x}
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