Re: why does push not default to $

That would break

push @a, @b;
[download]

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.

Comment on Re: why does push not default to $_? Select or Download Code

Replies are listed 'Best First'.
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;` [download] if @b is empty? It doesn't print $_ `for (1..13) { print @b; }` [download] Furthermore there is no error with: `push @a, @b;` [download] but you get a warning with `push @a;` [download] so perl can distinguish between missing parameters and empty parameters... Cheers Rolf	[reply] [d/l] [select]
Re^3: why does push not default to $_? by ikegami (Patriarch) on Dec 04, 2008 at 21:48 UTC
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?	[reply]
Re^4: why does push not default to $_? by LanX (Saint) on Dec 04, 2008 at 23:25 UTC
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	[reply] [d/l]
Re^5: why does push not default to $_? by ikegami (Patriarch) on Dec 05, 2008 at 02:36 UTC