in reply to Re^2: Why does foo() evaluate in array context in "${\foo()}"?
in thread Why does foo() evaluate in array context in "${\foo()}"?
because there is no such thing as a list in scalar context,
My take on it is that this is the exception that disproves that rule. Whilst others will disagree and probably decline to comment.
If you put that thing--two or more things separated by commas--that we're told is not a list, in a scalar context, at compile time, then the compiler knows the list is useless and can discard all but the last element. This is shown nicely by Deparse:
C:\test>perl -MO=Deparse -e"print scalar( 'fred', 'john', 'bill' )" print scalar('???', '???', 'bill'); -e syntax OK
And backed up by warnings:
C:\test>perl -we"print scalar( 'fred', 'john', 'bill' )" Useless use of a constant in void context at -e line 1. Useless use of a constant in void context at -e line 1. bill
But construct that same thing in a way that the context is not directly discernable at compile-time, and it has no choice but to construct a list er, leave the thing intact.
Again nicely shown by Deparse:
C:\test>perl -MO=Deparse -e"print scalar( sub{'fred', 'john', 'bill'}- +>() )" print scalar sub { 'fred', 'john', 'bill'; } ->(); -e syntax OK
And no warnings either:
C:\test>perl -we"print scalar( sub{'fred', 'john', 'bill'}->() )" bill
Just a list thing in a scalar context being reduce/resolved/resulted to the last element of that list thing.
what does \(LIST) become?
The only way left to describe it is that the thing (that would be a list in a list context), becomes the last element of that thing when it finds itself in a scalar context. :)
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Re^4: Why does foo() evaluate in array context in "${\foo()}"?
by ig (Vicar) on Apr 23, 2009 at 10:11 UTC |