Re: Next unused sequential number

Unless you have an array containing billions of uids, I'd do (untested):

my @uids = ....;   # List of used UIDs
my %uids; @uids{@uids} = ();
my $uid = 1001;
{$uid++, redo if exist $uids{$uid}}
say "First unused uid: $uid";
[download]

It doesn't actually need the list to be sorted.

Alternatively, you can make use of the fact the list is sorted (again, untested):

my @uids = ....;  # List of sorted uids
my $uid = 1001;
for (my $i = 0; $i < @uids; $i++) {
    next if $uids[$i] < $uid;
    last unless $uids[$i] == $uid;
    $uid++;
}
say "First unused uid: $uid";
[download]

Note that the latter algorithm isn't Perl specific, and can be trivially ported to many general purpose languages.

Comment on Re: Next unused sequential number Select or Download Code

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Re^2: Next unused sequential number by Anonymous Monk on Aug 05, 2011 at 18:17 UTC
Thank you for your collective wisdome perl monks. I found This to be quite helpful: `Alternatively, you can make use of the fact the list is sorted (again, + untested): my @uids = ....; # List of sorted uids my $uid = 1001; for (my $i = 0; $i < @uids; $i++) { next if $uids[$i] < $uid; last unless $uids[$i] == $uid; $uid++; } say "First unused uid: $uid";` [download] My day is going well now!	[reply] [d/l]

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Re^2: Next unused sequential number
by Anonymous Monk on Aug 05, 2011 at 18:17 UTC

Alternatively, you can make use of the fact the list is sorted (again,
+ untested): 

my @uids = ....;  # List of sorted uids
my $uid = 1001;
for (my $i = 0; $i < @uids; $i++) {
    next if $uids[$i] < $uid;
    last unless $uids[$i] == $uid;
    $uid++;
}
say "First unused uid: $uid";
[download]

[reply]
[d/l]