Re^3: array of binaries to hex conversion

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

How/what?

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

I don't think it does :) but I stick with pack

This ***might*** shed a little light

http://perl5.git.perl.org/perl.git?a=search&h=HEAD&st=grep&s=pp_oct

-> http://perl5.git.perl.org/perl.git/blob?f=pp.c#l2824

http://perl5.git.perl.org/perl.git?a=search&h=HEAD&st=grep&s=grok_bin

-> http://perl5.git.perl.org/perl.git/blob?f=numeric.c#l141

-> http://perldoc.perl.org/perlapi.html#grok_bin

Comment on Re^3: array of binaries to hex conversion

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Re^4: array of binaries to hex conversion by remiah (Hermit) on Sep 12, 2012 at 11:44 UTC
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.	[reply]

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Re^4: array of binaries to hex conversion
by remiah (Hermit) on Sep 12, 2012 at 11:44 UTC

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))));

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"

print  '0x'. unpack 'H*', pack("B32", substr("0" x 32 . join('',@rand1), -32));

[reply]