I need to make a simple conditional in a macro. If #1 is "1", "2", "3", "4", "5", or "6", then print Text: #1, otherwise, do nothing, even if it is empty. I tried this code:


I cannot even get make it able to check if 1 is present. How can I check #1 is equal to "1", "2", "3", "4", "5", or "6"?

  • Is #1 definitely an integer, or does the answer need to cover the case where it's not? Can it be an integer expression (such as 10 - 6) and if so does that need to count? – Joseph Wright Apr 22 '12 at 11:24
  • Only \macro{1}, \macro{2}, \macro{3}, \macro{4}, \macro{5}, and \macro{6} should count as a match. Something like \macro{15} should not count as any match. Usually, the macro is empty, like \macro{}, I guess is why my code is getting confused. – Village Apr 22 '12 at 11:29
  • Should \macro{\really\wild\input} be taken into account? – egreg Apr 22 '12 at 12:49

In your comments you indicated that #1 can be empty


would be an error in that case but a simple change to


will make an empty argument be Ok (and count as zero). beware that \macro{ 1} would then be an error as \ifnum0 1=1 does not work as 0 1 is not a valid number. It's possible to insert extra macros to remove spurious spaces if that is needed.

  • From the way the question is phrased, I'd interpret the comparison as essentially string-based (it has to be exactly as given). You could easily get rid of the spaces with \numexpr, or as you say using a macro-based approach. My worry with the \numexpr solution is that something like 1 + 1 would then be valid, but the question suggests it's not. – Joseph Wright Apr 22 '12 at 12:02

For a limited range of integers, with the other possibilities as blank or integer input, I would go with

\catcode`\@=11 %
  \ifnum 0#1 > 0 %
    \ifnum 0#1 < 7 %

used as


I hope that I understood your question correctly:

This is \mynumber{0} \mynumber{4} \mynumber{7}

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