Re^2: What is (a => b => c) ?

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Re^3: What is (a => b => c) ? by jeffa (Bishop) on Mar 27, 2015 at 16:12 UTC
Not if you declare/define it first: `no strict; sub c { print "see?\n" } @x = ( a => b => c );` [download] or `no strict; sub c; @x = ( a => b => c ); sub c { print "see?\n" }` [download] But not: `no strict; @x = ( a => b => c ); sub c { print "see?\n" }` [download] UPDATE: corrected 2nd example (used `&c;` which called the sub ... oops) jeffa L-LL-L--L-LL-L--L-LL-L-- -R--R-RR-R--R-RR-R--R-RR B--B--B--B--B--B--B--B-- H---H---H---H---H---H--- (the triplet paradiddle with high-hat)	[reply] [d/l] [select]
Re^4: What is (a => b => c) ? by Schmunzie (Beadle) on Mar 27, 2015 at 16:19 UTC
That's fascinating, amazing and scary in equal measure.	[reply]
Re^3: What is (a => b => c) ? by ikegami (Patriarch) on Mar 27, 2015 at 18:17 UTC
Not at all. Prepending `&` to a sub call causes the sub call to ignore the sub's prototype. Prepending `&` to a sub call with no parameter list specified (not even `()`) causes the sub call to ignore the sub's prototype, and forgoes the creation of a localized `@_`. These are not thing you should ever need to do. Because it must be followed by a sub call, `&` has the side effect of forcing the following word to be treated as the name of a sub, but that's just a side effect.	[reply] [d/l] [select]