Re: The [] metaoperator
Larry Wall skribis 2005-05-14 8:29 (-0700): > : say [x]; > : Is it a repeating metaoperator on an empty list, or a single-element > : array reference that contains the return value of calling &x()? > Always the first. [x] doesn't have to do lookahead. Does this mean that [x] is just an interesting way to create an empty list? And undef in scalar context? Very neat for obfuscation: return [%]; and let people wonder why the heck that [%] is there :) Juerd -- http://convolution.nl/maak_juerd_blij.html http://convolution.nl/make_juerd_happy.html http://convolution.nl/gajigu_juerd_n.html
Re: The [] metaoperator
On Sat, May 14, 2005 at 10:53:38PM +0800, Autrijus Tang wrote: : On Sat, May 14, 2005 at 10:56:29PM +1000, Damian Conway wrote: : > 8. To verify the monotonicity of a sequence: : > : >$is_monotonic = [<] @numbers; : : Hey. Does this mean that the [] metaoperator folds with the : associativity of the operator inside it? Yes. It's as if there is a long cat, only without the cat. : That is, if the operator inside is right-associative, it functions as : foldr; if the operator is left-associative, it functions as a foldl; and : if the operator is chain-associative like <, it assumes the special, : short-circuiting chained-folding semantic? : : [>] 1, 2, 3;# 1 > 2 > 3 # 3 is not evaluated : [**] 1, 2, 3;# 1 ** (2 ** 3) : [+] 1, 2, 3;# (1 + 2) + 3 : : If so, should I send patches to S03, or is it already in the works? Feel free, unless someone else volunteers. : Finally, what does this mean? : : say [x]; : : Is it a repeating metaoperator on an empty list, or a single-element : array reference that contains the return value of calling &x()? Always the first. [x] doesn't have to do lookahead. Larry
The [] metaoperator
On Sat, May 14, 2005 at 10:56:29PM +1000, Damian Conway wrote: > 8. To verify the monotonicity of a sequence: > >$is_monotonic = [<] @numbers; Hey. Does this mean that the [] metaoperator folds with the associativity of the operator inside it? That is, if the operator inside is right-associative, it functions as foldr; if the operator is left-associative, it functions as a foldl; and if the operator is chain-associative like <, it assumes the special, short-circuiting chained-folding semantic? [>] 1, 2, 3;# 1 > 2 > 3# 3 is not evaluated [**] 1, 2, 3;# 1 ** (2 ** 3) [+] 1, 2, 3;# (1 + 2) + 3 If so, should I send patches to S03, or is it already in the works? Finally, what does this mean? say [x]; Is it a repeating metaoperator on an empty list, or a single-element array reference that contains the return value of calling &x()? Thanks, /Autrijus/ pgpRdgO9r5PyG.pgp Description: PGP signature
