Range[1, n] /. n -> 3

produces an error message

 range::range: Range specification in Range[1,n] does not have appropriate bounds

even though it does produce correct output {1, 2, 3}.

What's a good way to accomplish the same thing while avoiding the error message?

  • 4
    $\begingroup$ ReplaceAll or /. does not hold its arguments, so Range[1, n] is evaluated with symbolic n first. $\endgroup$
    – ilian
    Sep 22 '15 at 19:29
  • $\begingroup$ Because it first evaluates Range[1,n] which is not very miningful and only then substitutes n->3 With[{n=3},Range[1,n]] could be more appropriate or R[n_]:=Range[1,n]; R[3] or as suggested by @ilian $\endgroup$ Sep 22 '15 at 19:32
  • $\begingroup$ Yet Attributes[ReplaceAll] gives no Hold attributes! $\endgroup$
    – murray
    Sep 22 '15 at 19:57
  • $\begingroup$ @murray that is exactly what illian says... $\endgroup$ Sep 22 '15 at 20:19
  • $\begingroup$ Sorry, I missed the "no" in illian's comment. All is clear now, thanks. $\endgroup$
    – murray
    Sep 22 '15 at 21:22

For your updated question I think you want to hold evaluation until all arguments are numeric.

myRange[x__?NumericQ] := Range[x]

myRange[1, n] /. n -> 3
{1, 2, 3}

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