Re: Questions on sort

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Re^2: Questions on sort by jiashiun (Initiate) on Aug 16, 2011 at 06:09 UTC
what is the purpose of putting the "[0]" in the second code? I got the same output with the "[0]" removed. `>perl -E "say('abc123def' =~ /(\d+)/);" 123` [download]	[reply] [d/l]
Re^3: Questions on sort by jwkrahn (Abbot) on Aug 16, 2011 at 06:56 UTC
In the code `($a =~ /(\d+)/)[0] <=> ($b =~ /(\d+)/)[0]` the regular expression is used in a list slice with a single index which returns a scalar value which can be compared with the `<=>` operator. Without the list slice the regular expression would return TRUE or FALSE in scalar context. In the code `my($aa) = $a =~ /(\d+)/;` the parentheses around `$aa` impose list context on the regular expression which in list context returns all the contents of capturing parentheses in the regular expression.	[reply] [d/l] [select]
Re^4: Questions on sort by locked_user sundialsvc4 (Abbot) on Aug 16, 2011 at 12:23 UTC
`++` ... Read the above post five times in a row.
Re^3: Questions on sort by ikegami (Patriarch) on Aug 16, 2011 at 06:49 UTC
The original code didn't use `say`, though. It used something that placed the expression scalar context, and there it makes a difference. `>perl -E"say(scalar( 'abc234def' =~ /(\d+)/ ));" 1 >perl -E"say(scalar( ('abc234def' =~ /(\d+)/)[0] ));" 234` [download]	[reply] [d/l] [select]
Re^3: Questions on sort by Anonymous Monk on Aug 16, 2011 at 06:27 UTC
Its the same purpose they would serve in an array `print( ( 4,6,8 )[0] )` [download]	[reply] [d/l]