Why is { } interpreted as an option in my function?

Question

I don't understand the output from:

Remove[g];
Options[g] = {"asd" -> 2};
g[a_, Optional[z_?Positive, 1], OptionsPattern[]] := {a, z, OptionValue["asd"]}

g[2, {}, "asd" -> 4]

{2, 1, 4}

The documentation of PatternTest says:

Any result for test[pval] other than True is taken to signify failure

but Positive[{}] =!= True holds, so it appears that the { } argument is interpreted as an option although the head is not Rule. I'm not sure how this should be fixed because I don't understand why it is wrong as it stands. What am I missing?

I am not sure I understand, {} fails Positive test so optional 1 is used, what is the problem? Sorry if I missed something — Kuba, Feb 01 '19 at 14:39
@m_goldberg - I feel like questions where the user misunderstands how Mathematica works are specifically on-topic here. For this question, closing as duplicate of Extra empty lists as function arguments might be appropriate, so that the next time someone gets confused about this and their search results bring them here, they get directed to a helpful answer. — Jason B., Feb 01 '19 at 14:58
Possible duplicate of Extra empty lists as function arguments — MarcoB, Feb 05 '19 at 04:45

score 5 · Accepted Answer · answered Feb 01 '19 at 15:45

O.k. from the comments and the OptionsPattern documentation:

Any nesting of empty lists will match OptionsPattern

I understand now why the pattern matched against my expectation. To make it stop and behave the way I originally had in mind, the last argument (OptionsPattern[]) must be replace by

e : OptionsPattern[] /; MatchQ[Unevaluated[e], Repeated[_Rule | _RuleDelayed]]

After realizing that nested lists are intended as a valid way to specify options, however, it makes more sense to just keep things as in the OP.

Why is { } interpreted as an option in my function?

1 Answers1