Re^3: How to find Unicode: 0x13 in File

[choroba]

> 0x01

Why do you specify the length in hex?

Also note that if you use a length greater than 1 (which you want to speed it up), you can find false positives: read $fh, my $char, 2 reports 0x13 present in the following file:

a1

because

$ perl -wE 'say unpack "H*", "a1"'
6131
 ~~

[james28909]

"Why is read length hex." - Because that is how I deal with most files I personally work with, so it is more of preference than anything.

The false positive does not show on my pc unless I use unpack('h*' $_); so I am sure it is a platform/architecture scenario between our PC's. Also, from the way I read the OP's question, I thought he is not looking for actually 0x13 in plaintext, I though he was looking for 0x13 after unpacking to hex. If I am wrong I apologize, I were actually hoping to learn something more than anything else.

I am not up to par on Unicode because i havent had to deal with it in any of the material I work on. So if I am way off in left field, I apologize.

[AnomalousMonk]

The false positive does not show on my pc unless I use unpack('h*' $_); so I am sure it is a platform/architecture scenario between our PC's.

Not a platform mismatch, I would say, but probably because you're still reading a single character at a time from the file. If more than one character is read, the  /13/ ambiguity | false positive can appear with either  'H*' or  'h*' unpack templates:

c:\@Work\Perl>perl -wMstrict -le
"print 'A: found 0x13!' if  unpack('H*',        'a1') =~ /13/;
 print 'B: found 0x13!' if  unpack('h*', qq{\x{1f}s}) =~ /13/;
"
A: found 0x13!
B: found 0x13!

(BTW: The  * in both  'H*' and  'h*' implies reading and operating on a string of more than one character.)

I am not up to par on Unicode because ...

... and because you value your sanity.

Update: Another, perhaps more general, code example:

c:\@Work\Perl>perl -wMstrict -le
"use Data::Dump qw(pp);
 ;;
 for my $s ('a1', qq{\x{1f}s}) {
   print q{'H*' found 0x13! in }, pp($s) if unpack('H*', $s) =~ /13/;
   print q{'h*' found 0x13! in }, pp($s) if unpack('h*', $s) =~ /13/;
   }
"
'H*' found 0x13! in "a1"
'h*' found 0x13! in "a1"
'h*' found 0x13! in "\37s"

[james28909]

I were able to reproduce that, However I have no idea why it does that. The following code seems to work as expected:

my ($one, $two) = unpack('(h2)*', 'a1');
#print "$one | $two\n";
print "found 0x13\n" if ($one =~ '13' | $two =~ '13');

But even doing this, it is splitting the two bytes into individual bytes. Is this behavior of unpack documented? I am using Active Perl version 5.16.3 as well.