in reply to why does push not default to $_?

That would break 

push @a, @b;

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It would push $_ when @b is empty instead of push nothing as expected.

Update: Or maybe not. That's doesn't break print, after all.

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Re^2: why does push not default to $_?
by LanX (Saint) on Dec 04, 2008 at 21:47 UTC
    Good argument, but whats the difference to 
     print @b;

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    if @b is empty? It doesn't print $_ 
    for (1..13) {
  print @b;
}

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    Furthermore there is no error with: 
    push @a, @b;

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    but you get a warning with 
    push @a;

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    so perl can distinguish between missing parameters and empty parameters...

    Cheers Rolf

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      Yeah, I already noticed and updated my node.

      The only possible reason I could see if an undue burden on the parser, since it would be the first function to default the second argument. You could send a perl (wishlist) bug report?

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        hmm, I thought it may be a compiler limitation on the first argument, but split defaults to $_ on its second argument, (' ' on first one)

        I think there is no logical reason!

        Cheers Rolf

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