http://www.perlmonks.org?node_id=992924


in reply to array of binaries to hex conversion

$ perl -le'
my @rand1 = qw( 0 1 0 1 0 1 0 1 0 1 1 1 0 1 1 0 0 0 1 );

sub stringdecimal {
    return unpack("N", pack("B32", substr("0" x 32 . shift, -32)));
    }

sub arraystring {
    my $string = join("", @_);
    return $string;
    }

my $bin = arraystring (@rand1);
my $dec = stringdecimal($bin);
my $hex = sprintf("0x%x", $dec);

print $hex;
'
0x2abb1

$ perl -le'
my @rand1 = qw( 0 1 0 1 0 1 0 1 0 1 1 1 0 1 1 0 0 0 1 );

my $hex = sprintf "0x%x", oct "0b" . join "", @rand1;

print $hex;
'
0x2abb1

[download]

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Re^2: array of binaries to hex conversion
by bebe (Novice) on Sep 11, 2012 at 06:57 UTC

    Perfect!! thanks a lot!!!!

[reply]
Re^2: array of binaries to hex conversion
by remiah (Hermit) on Sep 12, 2012 at 01:41 UTC

    Hello jwkrahn

    I have just read perlpacktut and I would like to have some clue for my 2 questions about this thread.

    What OP does is in short... 

    printf "0x%x\n",                #print it as hex
    unpack("N",                 #unpack it as 32bit int ,big endian
        pack("B32",             #pack it as MSB 32bit integer
            sprintf("%032s",    #32 length bit string
                join('',
                    @rand1))));

    [download]
    1. Does it really MSB ?

    MSB should pad '0' for right side, not left side, doesn't it ? 
    my @rand1 = qw( 1 );
    this currenty goes "00000000000000000000000000000001". Is it OK?

    2. How oct insures it is MSB and 32bit big endian?

    Does oct thinks it as 'long long' seeing the size of $Config{longlongsize}? How it thinks for MSB? 
    use Config;
print "int size=".$Config{intsize}."\n";
print "short size=".$Config{shortsize}."\n";
print "long size=".$Config{longsize}."\n";
print "long long size=".$Config{longlongsize}."\n";

    [download]
    If I could have some explanations by you or other monks, I am glad.

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[d/l]
[select]
[reply]

        Thanks for reply. Stories of bit is a little bit hard for me. Please point out if I am saying something wrong.

        How/what?

        I was wrong. I was confused with little endian and big endian. Here OP expects it's input bit array as Big Endian(N) and descending bit order(B). 

        printf "0x%x\n",                #hex
    unpack("N",                 #unpack it as 32bit int ,big endian
        pack("B32",             #pack it as MSB descending bit, 32bit inte
            sprintf("%032s",    #32 length bit string
                join('',
                    @rand1))));
        So, left pad works fine, even if it was an array of 16bit. 
        my $num;
$num=pack('n',7); #16bit short Big endian, it's value is 7
printf "big endian, descending bit order:%s\n", unpack('B16',$num);
printf "big endian, ascending  bit order:%s\n", unpack('b16',$num);
printf "unsigned short=%d\n",unpack('n',pack('B16', unpack('B16', $num)));

printf "right pad:%s\n", unpack('B16',$num) . '0' x 16;
printf "32bit right pad=%d\n",unpack('N',pack('B32', unpack('B16',$num) . '0' x 16));
#==> this prints "32bit right pad=458752"
printf "left  pad:%s\n", '0' x 16 . unpack('B16',$num);
printf "32bit left  pad=%d\n",unpack('N',pack('B32', sprintf('%032d', unpack('B16',$num)))
#==> this prints "32bit left  pad=7"
        You are the same anonymous with this. 
        print  '0x'. unpack 'H*', pack("B32", substr("0" x 32 . join('',@rand1), -32));
        So, you are going to say: If you need it's hexadecimal values, there is no need for decimal( unpack 'N') -> hexadecimal conversion( printf %x), just print it as hex( unpack 'H').

        I am going to see your links. Thanks for reply.

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