I think I have understood this correctly but please put me right if I am barking up the wrong tree

$_[0] & 0xff # A bitmask to gets the last (8 bit) byte from the intege
+r.  
             # It has exactly the same result as $_[0] % 256

>> 5         # bitshift the result of the above so all but
             # the 3 MSB drop off the end.

[download]
the return automagically maps this to the decimal for the three remaining bits. Was use integer necessary for this or does it improve efficiency ?

Here is a one liner to dump what is going on

perl -le'print join "\t", $_, (unpack "B*", pack "I*", $_),($_ & 0xFF)
+ >> 5 for (0..256)'

[download]
and some of the output for the interesting regions 
0       00000000000000000000000000000000        0
1       00000000000000000000000000000001        0
2       00000000000000000000000000000010        0
.
.
30      00000000000000000000000000011110        0
31      00000000000000000000000000011111        0
32      00000000000000000000000000100000        1
33      00000000000000000000000000100001        1
.
.
222     00000000000000000000000011011110        6
223     00000000000000000000000011011111        6
224     00000000000000000000000011100000        7
225     00000000000000000000000011100001        7
.
.
255     00000000000000000000000011111111        7
256     00000000000000000000000100000000        0

[download]

Cheers,
R.

Pereant, qui ante nos nostra dixerunt!

In reply to Re^2: calculating the 3 MSBs from an integer by Random_Walk
in thread calculating the 3 MSBs from an integer by insaniac

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