On 11/17/06, Rob Dixon <[EMAIL PROTECTED]> wrote:
Dharshana Eswaran wrote:
>
> On 11/16/06, Rob Dixon <[EMAIL PROTECTED]> wrote:
>>
>>>> Dharshana Eswaran wrote:
>>>>>
>>>>> I am working on the code below:
>>>>>
>>>>> my @hex = ("14", "2a", "11", "83");
>>>>> $hex_length = @hex;
>>>>>
>>>>> for($i=0; $i<$hex_length; $i++) {
>>>>>
>>>>> my $binary = unpack 'B*', pack 'H*', $hex[$i];
>>>>>
>>>>> print "$binary\n\n";
>>
>> Oh dear. It looks like you have an ealier version of Perl where unpack
has no
>> sub-templates, altthough I thought it had always been like that. Find
out
>> what version you are running (with perl -v) and consider upgrading. I
am
>> using v5.8.8 and bmost people use v5.8 or above.
>>
>> In the meantime, replace
>>
>> my @fields = unpack '(A7)*', $binary;
>>
>> with
>>
>> my @fields = $binary =~ /.{1,7}/g;
>
>
> my @fields = $binary =~ /.{7,1}/g; This works fine
>
> my @fields = $binary =~ /.{7,1}/g; Why does'nt this work?
>
> Coz i wanted the procedure to be reversed, that is if the input is
> "31","59", "4C","15","53"
> I wanted the output to be
>
> 0110001
> 0110010
> 0110001
> 0101010
> 1110001
> 01010
The input is "31","59", "4C","15","53","DD","54","31"
The above input is represented as shown below:
00110001 01011001 01001100 00010101 01010011 11011101 01010100 00110001
The correction in the desired output is
0110001
0110010
0110001
0101010
0110001
0101010
0110111
The output should have 7bits for each representation. That is 8th bit of the
1st byte is taken as the 0th bit of the next byte and 7th and 8th of the
second bytes becomes the 0th and 1st of the third byte, next 6th, 7th,8th of
the third becomes 0th,1st,2nd of the fourth....And so on....
The 1 and the 7 in the regex can't be exchanged as they are a minimum and
maximum number of characters to match. The pattern matches between one and
seven
of any character, and so splits the 40-character string into five blocks
of
seven characters and one of five.
I don't understand what you mean about reversing the procedure I'm afraid.
"31","59","4C","15","53"
is
00110001 01011001 01001100 00010101 01010011
in binary, and I can't see how to get to the bit patterns you give from
that,
although it starts off close enough. Your fourth pattern of seven bits,
for
instance, has three consecutive 1s in it, and that doesn't appear anywhere
in
the number sequence you gave.
Are you sure the data you've posted is right?
Rob
I hope the corrections would help you better.
Thanks and Regards,
Dharshana
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