Use of shift in Ternary Operator

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

Can someone explain to me why this is. Is the ternary operator some kind of list operator in disguise? Why does the context of shift need to be so explicit.

# WRONG! at least in 5.8.5 - context is ambiguous?
my $thing = shift ? 1: 0

#RIGHT!
$thing = shift() ? 1 : 0
[download]

if someone can point to the book that explains this, I would be greatful, otherwise it's a bug in my eyes because '?' is an operator and the interpretor upon hitting that operator as such and not a token should decipher the context of the shift.

Oct 05, 2025 at 20:56 UTC McDarren Moved from Snippets Section to SOPW

Comment on Use of shift in Ternary Operator Download Code

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Re: Use of shift in Ternary Operator by hipowls (Curate) on Mar 22, 2008 at 23:00 UTC
The problem is that `?` is ambiguous, is it part of a ternery operator or the start of a `??` match? From Perl Regular Expressions Reference `?pattern?` is like `m/pattern/` but matches only once. No alternate delimiters can be used. Must be reset with `reset()`. puting `()` after `shift` resolves the ambiguity. Update:To explain further, `shift` expects a single argument which is an expression that evaluates to an array. If no argument is provided it defaults to `@_`. When perl sees `shift ?` it doesn't know if you mean `shift(@_) ? 1: 2;` [download] or something like `shift ?pattern? ? @a: @b;` [download] the parentheses forces the first interpretation and shift defaults to using `@_`.	[reply] [d/l] [select]
Re: Use of shift in Ternary Operator by jwkrahn (Abbot) on Mar 22, 2008 at 19:38 UTC
Is the ternary operator some kind of list operator in disguise? Yes, the Conditional Operator can process either scalars or lists. Warning: Use of "%s" without parentheses is ambiguous	[reply]