Re^3: Local for lexicals

Seems you just want a complicated way of not writing $_[0] in the sub. That can be done of course. But why would you? And what has this to do with localizing a lexical, which you were asking about in your earlier posts? If you're passing in the parameter to the sub, there's no need to localize anything, is there?

my $f = sub {my $x = shift; sub {$x ** 2}->()};
say $f->(2);  # => 4
say $f->(3);  # => 9
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Re^4: Local for lexicals by LanX (Saint) on Aug 11, 2009 at 00:11 UTC
because $_[0] is an alias and $x is only a copy, he wants $x to be an alias of the first arg! see explanation of second code example in Re^3: Local for lexicals JadeNB should really post better test cases to avoid confusion! Cheers Rolf	[reply]
Re^4: Local for lexicals by JadeNB (Chaplain) on Aug 12, 2009 at 19:26 UTC
Seems you just want a complicated way of not writing `$_[0]` in the sub. That can be done of course. But why would you? And what has this to do with localizing a lexical, which you were asking about in your earlier posts? If you're passing in the parameter to the sub, there's no need to localize anything, is there? In order: Yes, that's right (for the `lambda` example). Skipped. Why not? Only that `local`ising a lexical was my first thought for how to implement it. (Example code is at Local for lexicals.) Once I thought of the problem in that context, it immediately seemed to crop up many other places. You are certainly right that there are ways to do it without `local`isation. If I write out the sub—as you've done here: `my $f = sub {my $x = shift; sub {$x 2}->()};` [download] (formatted, I think, specifically to be faithful to my original attempt), or even more briefly as just: `my $f = sub { my $x = shift; $x2 }` [download] then it certainly works. I would just like to avoid the slight extra verbiage and write instead `my $f = lambda $x => sub { $x**2 }` [download] as syntactic sugar (well, for some definition of ‘sugar’) for the same thing.	[reply] [d/l] [select]