Hi monks,

I'm using Perl to parse the byte stream output of a hardware.
Depending on the HW configuration I get a stream of data consisting
of 40, 48, 56 or 64 bits litle-endian.
In principle the lower 5,6,7 or the complete 8 byte of a 64bit litle-endian integer.
I was trying to convert this data using unpack but whatever I tried using 'x' or '@'
I did not succeed in (p)adding the missing 0 bytes before converting to a 64bit integer.
(I'm using a Perl with support for 64bit integers)

My current solution looks like: (using bitstrings..)
my $bytes_per_value = 5;

# to simulate the byte stream  using 40bit = 5 * 8bit;
my $value = 0xf_dead_beef_4;
my $bin_value = substr (pack ('Q', $value), 0, $bytes_per_value);
my $buffer = $bin_value x 4

my $nbytes = length ($buffer);

my $fmt = sprintf "(b%d)*", $bytes_per_value << 3;
my @stream_data = unpack ($fmt, substr ($buffer, 0, $nbytes));
my @values = map { oct '0b'.reverse ($_)} @stream_data;
foreach my $v (@values) {
   printf "0x%x\n", $v;
}

[download]

My question:
Is there a way to unpack this stream directly into an array of QWords (64bit)
using some form of unpack for 5,6 and 7 byte data. (The 8 byte case is obviously easy :-) )

Means: Can I specify via the format string to convert 5,6 or 7 bytes + 3,2 or 1 padding Zero bytes
to a 64bit integer?
Or can unpack only work on "existing" data bytes.

Thanks for any hints!

Axel

In reply to Can unpack add zero bytes before converting? by mossi2000

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