parsing 'length ...'

rsFalse has asked for the wisdom of the Perl Monks concerning the following question:

Hello. I tried to avoid some parenthesis and have got some problems.
Firstly, I wanted to write this code in fashion of ternary operator:

perl -le '$_ = "a"; print do { if( length ){ "A" } else{ "B" } };'
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OUTPUT:

A
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There is with "ternary":

perl -le '$_ = "a"; print length ? "A" : "B" '
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OUTPUT:

Warning: Use of "length" without parentheses is ambiguous at -e line 1
+.
Use of ?PATTERN? without explicit operator is deprecated at -e line 1.
Search pattern not terminated or ternary operator parsed as search pat
+tern at -e line 1.
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I forgot that 'length' is unary operator with higher precedence, so it interpret following '?' as matching parens. Same if I would write: 'print length // "A" ' - error, when 'print length || "A" ' - is ok.
So I tried to avoid 'length' to interpret '?' as regex in this way:

perl -le '$_ = "a"; print length . '' ? "A" : "B" '
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But it still outputs:

Use of ?PATTERN? without explicit operator is deprecated at -e line 1.
Search pattern not terminated or ternary operator parsed as search pat
+tern at -e line 1.
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Concatenation point has higher precedence than unary 'length' so I think 'length' should give its value which will be concatenated with empty line, and later ternaty op., and later - 'print'.
If I use '+ 0' instead, I got warning of ambiguity (but '+' can't start a regex, and for me seems that there is no ambiguity):

perl -le '$_ = "a"; print length + 0 ? "A" : "B" '
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Warning: Use of "length" without parentheses is ambiguous at -e line 1
+.
A
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Re: parsing 'length ...'
by Athanasius (Archbishop) on Feb 24, 2016 at 15:38 UTC

Hello rsFalse,

perl -le '$_ = "a"; print length . '' ? "A" : "B" '
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The one-liner terminates at the second '. Try using "" for the empty string instead of '' here:

 1:34 >perl -le "$_ = 'a'; print length . '' ? 'A' : 'B'"
A

 1:34 >
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(but my single and double quotes are reversed because I’m on Windows.)

Hope that helps,

Athanasius <°(((>< contra mundum Iustus alius egestas vitae, eros Piratica,

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Re: parsing 'length ...'
by Anonymous Monk on Feb 24, 2016 at 16:19 UTC

Warning: Use of "length" without parentheses is ambiguous
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$ perl -le '$_ ="a"; print length() ? "A" : "B"'
A
$
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Re: parsing 'length ...'
by kcott (Archbishop) on Feb 25, 2016 at 06:17 UTC

G'day rsFalse,

A couple of tips from someone who's fallen foul of both the problems you've encountered on more than one occasion:

Whenever I get a "Use of "whatever" without parentheses is ambiguous" warning, adding the parentheses, rather than casting around for some other workaround, is generally the easiest solution.
Quotes within one-liners are problematic. Consider using the various q{}, qq{}, etc. Quote-Like Operators instead of literal quotation marks.

— Ken

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