first 5 bits and last 3 bits

pashanoid has asked for the wisdom of the Perl Monks concerning the following question:

I've never dealt with bits and bytes, so I'm unable to figure out the first 5 bits and last 3 bits of say 129... supposed to get 16 and 3 or 2 or something... Thank you!

I sort of came up with this, but it's not nice:

#!/usr/bin/perl
#

$num = 129;

$str = unpack("B32", pack("N", $num));
$str =~ s/^0+(?=\d)//;   # otherwise you'll get leading zeros
print "$str\n";

$fragment =  substr $str, 0, 5; #first 5 bytes
print "$fragment\n";
$num1=unpack("N", pack("B32", substr("0" x 32 . $fragment, -32)));
print "num1=$num1\n";

$fragment =  substr $str, 5, 3; #first 5 bytes
print "$fragment\n";
$num2=unpack("N", pack("B32", substr("0" x 32 . $fragment, -32)));
print "num2=$num2\n";
[download]

Comment on first 5 bits and last 3 bits Download Code

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Re: first 5 bits and last 3 bits by moritz (Cardinal) on Jul 26, 2011 at 12:39 UTC
The "first 5 bits" of a number really depend on the storage format. For example as a 32 bit integer, 129 is represented as `00000000000000000000000010000001`, the first 5 bits of which are all zero. If that's not what you want, consider converting the number to the string `10000001` by using sprintf and then use substr to extract what you want. Perl 6 - second systems done right	[reply] [d/l] [select]
Re^2: first 5 bits and last 3 bits by pashanoid (Scribe) on Jul 26, 2011 at 12:55 UTC
It's 8 bits	[reply]
Re^3: first 5 bits and last 3 bits by cdarke (Prior) on Jul 26, 2011 at 13:08 UTC
Not just how many bits, but whether it is big or little-endian.	[reply]
Re: first 5 bits and last 3 bits by fisher (Priest) on Jul 26, 2011 at 12:38 UTC
Use bitmask and bitwise boolean operations. `fisher% perl -e 'my $byte=129; my $first5=$byte&0xf8; printf "%08b, %d +\n", $byte, $first5;' 10000001, 128` [download]	[reply] [d/l]
Re^2: first 5 bits and last 3 bits by DrHyde (Prior) on Jul 27, 2011 at 09:26 UTC
If you're creating a bitmask, it's clearer to say `0b11111000` than `0xf8`.	[reply] [d/l] [select]
Re: first 5 bits and last 3 bits by ikegami (Patriarch) on Jul 26, 2011 at 18:55 UTC
`my $byte = 129; # 0b10000001 = 129 my $most_significant_5 = $byte >> 3; # 0b10000 = 16 my $least_significant_3 = $byte & 0x7; # 0b 001 = 1` [download] How did I arrive at that? Basically, you want to either Mask out the bits of higher significance than those you want, then shift the bits you want to index 0; or Shift the bits you want to index 0 then mask out the bits of higher significance than those you want. `is a simplification of ---------------------------------------------------- $byte >> 3 ( $byte & 0xFF ) >> 3 or ( $byte >> 3 ) & 0x1F $byte & 0x7 ( $byte & 0x7 ) >> 0 or ( $byte >> 0 ) & 0x7` [download] Notes; «`0x7`» is `0b111` or 2³-1. «`& 0x7`» "keeps" only the least significant 3 bits. «`0x1F`» is `0b11111` or 2⁵-1. «`& 0x7`» "keeps" only the least significant 5 bits. «`0xFF`» is `0b11111111` or 2⁸-1. «`& 0x7`» "keeps" only the least significant 8 bits. «`& 0xFF`» is a no-op when dealing with 8-bit values. «`>> 0`» is a no-op for integers, which is what we're dealing with.	[reply] [d/l] [select]
Re^2: first 5 bits and last 3 bits by pashanoid (Scribe) on Jul 27, 2011 at 06:15 UTC
thank you so much for the answer and education!!!	[reply]
Re: first 5 bits and last 3 bits by jpl (Monk) on Jul 26, 2011 at 20:32 UTC
This posting may win some sort of award for ambiguity. What do "first" and "last" mean? Are the "first" the most significant, and the "last" the least significant? If, as the OP added, "it's 8 bits", which 8 bits? The most/least significant? If you strip away leading 0's, as the OP did, is it the 8 bits whose most significant bit is a 1 (which, of course, may not yield 8 bits)? There are many reasonable suggestions here, just as there are many reasonable interpretations of what the heck the OP wants. It doesn't help that the OP offered several suggestions on what the result should be.	[reply]
Re^2: first 5 bits and last 3 bits by ikegami (Patriarch) on Jul 26, 2011 at 21:30 UTC
It doesn't help that the OP offered several suggestions on what the result should be. He seems sure of 16, which is indeed the result one gets from taking the most significant 5 bits of an 8 bit number. As far as I'm concerned, this clears up the ambiguity, leaving a post that's simply unclear.	[reply]

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