Re: why does push not default to $

Well, I think print and push are different in that when you see:

  print;
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it feels like you want to print something, that thing being directly related to the context (default arg) makes sense. on the other hand:

  push @a;
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feels like we want to push @a to some place... like a stack. If there was a default stack than the right action would be to push @a to the default stack. So, I guess it is more about linguistics than implementation,

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Re^2: why does push not default to $_? by plobsing (Friar) on Dec 06, 2008 at 09:35 UTC
There is a default stack. At least according to `shift` and `pop`. Making 1-arg `push` (and `unshift`) work with `@_` (or `@ARGV`) as a default 1st arg would, IMHO, be more consistent (I have accidentally expected this behaviour once). Unfortunately, its not terribly useful. I don't find myself adding to `@_` frequently.	[reply] [d/l] [select]
Re^3: why does push not default to $_? by ikegami (Patriarch) on Dec 07, 2008 at 17:47 UTC
It's not a good idea. The only thing you could push onto the "stack" is would be the contents of an array. For example, you can't make `push $foo;` work without breaking `push @foo, $x;`. The usefulness of the feature is just too limited to compensate for the confusion the its implementation would cause.	[reply] [d/l] [select]