FW: [Dub-pm] Quick pack question.

[Mcnamara John]

> How do I take a string formated like so, bitwise-speaking;
>
> 11111112 22222233 33333444 55556666 ...
>
> (that is, the first seven bits of the first octet correspond to the
> first character, the last bit of the first octet concatenated with
> the first six of the next correspond to the second character ...)


Or without pack:

    #!/usr/bin/perl -wl

    my $str = sprintf "%032b", 123456789;
    print $str, "\n";


    my $offset = 0;

    while ($offset < length $str) {
        my $septet = oct 'b' . substr($str, $offset, 7);

        printf "%07b %d\n", ($septet) x 2;
        $offset += 7;
    }


    __END__
    Prints:

    00000111010110111100110100010101

    0000011 3
    1010110 86
    1111001 121
    1010001 81
    0000101 5


This handles the trailing bits less than 7 differently from the other
example. Bug or feature, you decide. ;-)

John.
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