Re^2: Compare two arrays of simple numbers

return 0 if @$a != @$b;

Well that's (worst case) O(n) right there. :-)
Does != even have a meaning in list context ?

Update: Ouch, I'm so wrong it's ridiculous.

-David

Comment on Re^2: Compare two arrays of simple numbers Select or Download Code

Replies are listed 'Best First'.
Re^3: Compare two arrays of simple numbers by Prof Vince (Friar) on Oct 03, 2007 at 08:36 UTC
Well that's (worst case) O(n) right there. :-) That's what linear means. Does != even have a meaning in list context ? Edit: It doesn't force scalar context, but seems to compare array lengths. See : `sub zero { return (0, 0) } print +(zero() == 2) ? "scalar\n" : "list\n"; my @a = (0, 1); my @b = (1, 0); my @c = (0, 2 , 1); print +(@a == @b) ? "same length\n" : "not\n"; print +(@a == @c) ? "same length\n" : "not\n";` [download] Anyway, I think perl doesn't actually build the lists and directly optimize that to the array size, which should make it constant time.	[reply] [d/l]

Replies are listed 'Best First'.

Re^3: Compare two arrays of simple numbers
by Prof Vince (Friar) on Oct 03, 2007 at 08:36 UTC

Well that's (worst case) O(n) right there. :-)

Does != even have a meaning in list context ?

sub zero { return (0, 0) }
print +(zero() == 2) ? "scalar\n" : "list\n";
my @a = (0, 1);
my @b = (1, 0);
my @c = (0, 2 , 1);
print +(@a == @b) ? "same length\n" : "not\n";
print +(@a == @c) ? "same length\n" : "not\n";
[download]

[reply]
[d/l]