http://www.perlmonks.org?node_id=1017270


in reply to Re^2: Golf: reverse sort /etc/passwd by UID
in thread Golf: reverse sort /etc/passwd by UID

D'oh! I did omit a couple of bits didn't I (It was (obviously) untested -- I don't have an /etc/passwd).

I tried to remove the $b=~ & $a=~ bits when I noticed that reverse was one character less -- ignoring/forgetting that $_ cannot be both at the same time :(.

I don't think the (...)[0] bits are necessary, without /g only the regex will only return one (the first) match.

This should work? (still untested):

perl -E"say sort{$b=~/:(\d+):/<=>$a=~/:(\d+):/}<>" /etc/passwd

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Re^4: Golf: reverse sort /etc/passwd by UID
by thundergnat (Deacon) on Feb 05, 2013 at 20:33 UTC
    No, unfortunately that doesn't work either. A regex match in scalar context returns 1 or 0. It needs to be called in list context to return the value. 
    >perl -E"print 'foo'=~/(o+)/;"
oo

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    >perl -E"print scalar 'foo'=~/(o+)/;"
1

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Re^4: Golf: reverse sort /etc/passwd by UID
by Tommy (Chaplain) on Feb 05, 2013 at 20:26 UTC

    No, no. It does work (tested and confirmed). See my reply to thundergnat.

    NOTE- remember you have to omit the typos though, as I already pointed out. So far, the winner remains:

    UPDATE: thundergnat is right that the code below DOESN'T work as intended

    perl -e 'print reverse sort{/(:\d+:)/<=>/(:\d+:)/}<>' /etc/passwd

    Tommy
    A mistake can be valuable or costly, depending on how faithfully you pursue correction
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      No, no. It doesn't work (tested and confirmed).

      The sort block operates on the global variables $a and $b. A match without an explicit variable operates on $_. There is nothing mapping $a and $b to $_ there, so the sort block is just comparing 0 to 0 and leaving the order alone. The fact that /etc/password is in a "sorted" order makes it SEEM like it is working... but it isn't.

      print reverse sort{/(:\d+:)/<=>/(:\d+:)/}<DATA>;

__DATA__

aaa:5:
bbb:3:
ccc:1:
ddd:7:
eee:8:

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      eee:8:
ddd:7:
ccc:1:
bbb:3:
aaa:5:
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