Re: cut of first char of a string

Well, here's a couple of simple ones:

print join'',@{[do{shift@{$s=[split//,$s='foobarbaz']};@$s}]} 

print $s=reverse@{$s=do{pop @{$s=[split//,reverse$s='foobarbaz']};$s}}
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Although, I'm honestly not sure why the $s= that immediately precedes the do is necessary in the second one. Can anyone explain it to me?

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Re^2: cut of first char of a string by ambrus (Abbot) on Jul 04, 2005 at 08:47 UTC
I'm not sure but I think it's needed because you assign an anonymous array to `$s` before it, so if you do not set `$s`, the rightmost `$s` will give that array.	[reply] [d/l] [select]