thanos1983:

Questions about manipulating binary numbers like that are often best solved by looking them over in binary. Using a little bit of perl:

#!/usr/bin/perl
use strict;
use warnings;

my @memory = (0, 0, 5, 59);

print "Original values:\n";
printf "% 3u : %08b\n", $_, $_ for @memory;

$memory[1] = $memory[1] << 7;
$memory[2] = $memory[2] << 14;
$memory[3] = $memory[3] << 21;
print "\nAfter shifting the values\n";
printf "% 10u : %032b\n", $_, $_ for @memory;

my $t = 0;
$t += $_ for @memory;
print "\nThe sum:\n";
printf "% 10u : %032b\n", $t, $t;
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Running it gives a hopefully familiar number: ;^D

Original values:
  0 : 00000000
  0 : 00000000
  5 : 00000101
 59 : 00111011

After shifting the values
         0 : 00000000000000000000000000000000
         0 : 00000000000000000000000000000000
     81920 : 00000000000000010100000000000000
 123731968 : 00000111011000000000000000000000

The sum:
 123813888 : 00000111011000010100000000000000
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Update: I just read the thread, and noticed that in your code, you use the same construct several times. Additionally, you mention that you only want seven bits. So I'd suggest writing a subroutine that masks off the high bit of your bytes and packs the number for you. Something like the following, but with a better name:

sub plonk {
    my ($b0, $b1, $b2, $b3) = @_;
    return     $b0 & 0x7f
            + ($b1 & 0x7f) << 7
            + ($b2 & 0x7f) << 14
            + ($b3 & 0x7f) << 21;
}
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...roboticus

When your only tool is a hammer, all problems look like your thumb.