How can I generate non-repetitive random 4 bytes hex values in Perl?

Question

I want to generate random hex values and those values should not be repetitive and it should be of 4 bytes (ie: 0x00000000 to 0xffffffff) and the display output should contain leading zeros.

For example: if I get the value 1 it should not represented as 0x1 but 0x00000001.

I want a minimum of 100 random values. Please tell me: how can I do that in Perl?

Answer 1

To get a random integer:

int(rand(0x10000000))

To format it as 8 hexadecimal digits:

printf "%08x", int(rand(0x10000000))

Answer 2

perl -e 'printf "%08X\n", int rand 0xFFFFFFFF for 1 .. 100'

Answer 3

use LWP::Simple "get";
use List::MoreUtils "uniq";
print for uniq map { s/\t//, "0x$_" } split /^/, LWP::Simple::get('http://www.random.org/integers/?num=220&min=0&max=65535&col=2&base=16&format=plain&rnd=date.2009-12-14');

Adjust the url (see the form on http://www.random.org/integers/?mode=advanced) to not always return the same list. There is a minuscule chance of not returning at least 100 results.

Note that this answer is intentionally "poor" as a comment on the poor question. It's not a single question, it's a bunch all wrapped up together, all of which I'd bet have existing answers already (how do I generate a random number in range x, how do I format a number as a hex string with 0x and 0-padding, how do I add only unique values into a list, etc.). It's like asking "How do I write a webserver in Perl?" Without guessing what part the questioner really wants an answer to, you either have to write a tome for a response, or say something like:

perl -MIO::All -e 'io(":80")->fork->accept->(sub { $_[0] < io(-x $1 ? "./$1 |" : $1) if /^GET \/(.*) / })'

Answer 4

To get a random number in the range 0 .. (2<<32)-1:

my $rand = int(rand(0x100000000));

To print it in hex with leading zeroes:

printf "%08x", $rand;

Do please note this from the Perl man page:

Note: If your rand function consistently returns numbers that are too large or too small, then your version of Perl was probably compiled with the wrong number of RANDBITS

If that's a concern, do this instead:

printf "%04x%04x", int(rand(0x10000)), int(rand(0x10000));

Note, also, that this does nothing to prevent repetition, although to be honest the chance of a repeating 32 bit number in a 100 number sequence is pretty small.

If it's absolutely essential that you don't repeat, do something like this:

my (%a);                       # create a hash table for remembering values
foreach (0 .. 99) {
    my $r;
    do {
        $r = int(rand(0x100000000));
    } until (!exists($a{$r})); # loop until the value is not found
    printf "%08x\n", $r;       # print the value
    $a{$r}++;                  # remember that we saw it!
}

For what it's worth, this algorithm shouldn't be used if the range of possible values is less than (or even near to) the number of values required. That's because the random number generator loop will just repeatedly pull out numbers that were already seen.

However in this case where the possible range is so high (2^32) and the number of values wanted so low it'll work perfectly. Indeed with a range this high it's about the only practical algorithm.

Answer 5

Alnitak explained it, but here's a much simpler implementation. I'm not sure how everyone starting reaching for do {} while since that's a really odd choice:

my $max = 0xFFFF_FFFF;

my( %Seen, @numbers );

foreach ( 1 .. 100 )
 {
 my $rand = int rand( $max + 1 );
 redo if $Seen{$rand}++;
 push @numbers, $rand;
 }

print join "\n", map { sprintf "0x%08x", $_ } @numbers;

Also, as Alnitak pointed out, if you are generating a lot of numbers, that redo might cycle many, many times.

These will only be pseudorandom numbers, but you're not really asking for real random number anyway. That would involve possible repetition. :)