I've been having this problem in Perl for a few days now, and after scouring countless man pages, perldocs and googling too many search terms, hopefully someone here can help me out.
I am given two strings which represent hex values, i.e. "FFFF", not the Perl hex number 0xFFFF. Given two of these strings, I wish to convert them to binary form, perform a bitwise AND of the two, then take the output of this and examine each bit from LSB to MSB.
I have two problems right now; converting the hex string into a hex number, and shifting
the result of the bitwise AND.
For converting the hex string into a hex number, I've tried the following approaches which don't seem to work when I print them out to examine:
$a = unpack("H*", pack("N*", $a));
$a = sprintf("%H", $a);
Using a 'print' to examine each of these does not show a correct value, nor does using 'sprintf' either...
The second problem I have occurs after I perform a bitwise AND, and I want to examine each bit by shifting right by 1. To avoid the previous problem, I used actual Perl hex numbers instead of hex strings (0xffff instead of "ffff"). If I try to perform a shift right as follows:
#Convert from hex number to binary number
$a = sprintf("%B", $a);
$b = sprintf("%B", $b);
$temp = pack("B*", $a) & pack("B*", $b);
$output = unpack("B*", $temp);
At this point everything looks fine, and using a 'print' I can see that the values of the AND operation look right, but when I try to shift as follows:
$output = pack("B*", $output);
$output = $output >> 1;
$output = unpack("B*", $output);
The resulting value I get is in binary form but not correct.
What is the correct way of performing this kind of operation?
There's no such thing as a "hex number". A number is a number, a hexadecimal representation of a number is just that - a representation.
Just turn it into a number and use bitwise and.
my $num = (hex $a) & (hex $b);
print ($num & 1, "\n") while ($num >>= 1)
Related
I would like to normalize the variable from ie. 00000000.1, to 0.1 using Perl
my $number = 000000.1;
$number =\~ s/^0+(\.\d+)/0$1/;
Is there any other solution to normalize floats lower than 1 by removing upfront zeros than using regex?
When I try to put those kind of numbers into an example function below
test(00000000.1, 0000000.025);
sub test {
my ($a, $b) = #_;
print $a, "\n";
print $b, "\n";
print $a + $b, "\n";
}
I get
01
021
22
which is not what is expected.
A number with leading zeros is interpreted as octal, e.g. 000000.1 is 01. I presume you have a string as input, e.g. my $number = "000000.1". With this your regex is:
my $number = "000000.1";
$number =~ s/^0+(?=0\.\d+)//;
print $number;
Output:
0.1
Explanation of regex:
^0+ -- 1+ 0 digits
(?=0\.\d+) -- positive lookahead for 0. followed by digits
Learn more about regex: https://twiki.org/cgi-bin/view/Codev/TWikiPresentation2018x10x14Regex
Simplest way, force it to be treated as a number and it will drop the leading zeros since they are meaningless for decimal numbers
my $str = '000.1';
...
my $num = 0 + $str;
An example,† to run from the command-line:
perl -wE'$n = shift; $n = 0 + $n; say $n' 000.1
Prints 0.1
Another, more "proper" way is to format that string ('000.1' and such) using sprintf. Then you do need to make a choice about precision, but that is often a good idea anyway
my $num = sprintf "%f", $str; # default precision
Or, if you know how many decimal places you want to keep
my $num = sprintf "%.3f", $str;
† The example in the question is really invalid. An unquoted string of digits which starts with a zero (077, rather than '077') would be treated as an octal number except that the decimal point (in 000.1) renders that moot as octals can't be fractional; so, Perl being Perl, it is tortured into a number somehow, but possibly yielding unintended values.
I am not sure how one could get an actual input like that. If 000.1 is read from a file or from the command-line or from STDIN ... it will be a string, an equivalent of assigning '000.1'
See Scalar value constructors in perldata, and for far more detail, perlnumber.
As others have noted, in Perl, leading zeros produce octal numbers; 10 is just a decimal number ten but 010 is equal to decimal eight. So yeah, the numbers should be in quotes for the problem to make any sense.
But the other answers don’t explain why the printed results look funny. Contrary to Peter Thoeny’s comment and zdim’s answer, there is nothing ‘invalid’ about the numbers. True, octals can’t be floating point, but Perl does not strip the . to turn 0000000.025 into 025. What happens is this:
Perl reads the run of zeros and recognises it as an octal number.
Perl reads the dot and parses it as the concatenation operator.
Perl reads 025 and again recognises it as an octal number.
Perl coerces the operands to strings, i.e. the decimal value of the numbers in string form; 0000000 is, of course, '0' and 025 is '21'.
Perl concatenates the two strings and returns the result, i.e. '021'.
And without error.
(As an exercise, you can check something like 010.025 which, for the same reason, turns into '821'.)
This is why $a and $b are each printed with a leading zero. Also note that, to evaluate $a + $b, Perl coerces the strings to numbers, but since leading zeros in strings do not produce octals, '01' + '021' is the same as '1' + '21', returning 22.
I am new to Perl and I have difficulties using the different types.
I am trying to get an hexadecimal register, transform it to binary, use it a string and get substrings from the binary string.
I have done a few searches and what I tried is :
my $hex = 0xFA1F;
print "$hex\n";
result was "64031" . First surprise : can't I print the hex value in Perl and not just the decimal value ?
$hex = hex($hex);
print "$hex\n";
Result was 409649. Second surprise : I would expect the result to be also 64031 since "hex" converts hexadecimal to decimal.
my $bin = printf("%b", $hex);
It prints the binary value. Is there a way to transform the hex to bin without printing it ?
Thanks,
SLP
Decimal, binary, and hexadecimal are all text representations of a number (i.e. ways of writing a number). Computers can't deal with these as numbers.
my $num = 0xFA1F; stores the specified number (sixty-four thousand and thirty-one) into $num. It's stored in a format the hardware understands, but that's not very important. What's important is that it's stored as a number, not text.
When print is asked to print a number, it prints it out in decimal (or scientific notation if large/small enough). It has no idea how the number of created (from a hex constant? from addition? etc), so it can't determine how to output the number based on that.
To print an number as hex, you can use
my $hex = 'FA1F'; # $hex contains the hex representation of the number.
print $hex; # Prints the hex representation of the number.
or
my $num = 0xFA1F; # $num contains the number.
printf "%X", $num; # Prints the hex representation of the number.
You are assigning a integer value using hexadecimal format. print by default prints numbers in decimal format, so you are getting 64031.
You can verify this using the printf() by giving different formats.
$ perl -e ' my $num = 0xFA1F; printf("%d %X %b\n", ($num) x 3 ) '
64031 FA1F 1111101000011111
$ perl -e ' my $num = 64031; printf("%d %X %b\n", ($num) x 3 ) '
64031 FA1F 1111101000011111
$ perl -e ' my $num = 0b1111101000011111; printf("%d %X %b\n", ($num) x 3 ) '
64031 FA1F 1111101000011111
$
To get the binary format of 0xFA1F in string, you can use sprintf()
$ perl -e ' my $hex = 0xFA1F; my $bin=sprintf("%b",$hex) ; print "$bin\n" '
1111101000011111
$
lets take each bit of confusion in order
my $hex = 0xFA1F;
This stores a hex constant in $hex, but Perl doesn't have a hex data type so although you can write hex constants, and binary and octal constants for that matter, Perl converts them all to decimal. Note that there is a big difference between
my $hex = 0xFA1F;
and
my $hex = '0xFA1F';
The first stores a number into $hex, which when you print it out you get a decimal number, the second stores a string which when printed out will give 0xFAF1 but can be passed to the hex() function to be converted to decimal.
$hex = hex($hex);
The hex function converts a string as if it was a hex number and returns the decimal value and, as up to this point, $hex has only ever been used as a number Perl will first stringify $hex then pass the string to the hex() function to convert that value from hex to decimal.
So to the solution. You are almost there with printf(),there is a function called sprintf() which takes the same parameters as printf() but instead of printing the formatted value returns it as a string. So what you need is.
my $hex = 0xFA1F;
my $bin = sprintf("%b", $hex);
print $bin;
Technical note:
Yes I know that Perl stores all its numbers internally as binary, but lets not go there for this answer, OK?
If you're ok with using a distribution, I wrote Bit::Manip to make my prototyping a bit easier when dealing with registers (There's also a Pure Perl version available if you have problems compiling the XS code).
Not only can it fetch out bits from a number, it can toggle, clear, set etc:
use warnings;
use strict;
use Bit::Manip qw(:all);
my $register = 0xFA1F;
# fetch the bits from register using msb, lsb
my $msbyte = bit_get($register, 15, 8);
print "value: $msbyte\n";
print "bin: " . bit_bin($msbyte) . "\n";
# or simply:
# printf "bin: %b\n", $msbyte;
Output:
value: 250
bin: 11111010
Here's a blog post I wrote that shows how to use some of the software's functionality with an example datasheet register.
I am fresher in writing scripts so need some help.
so the issue is in one of my script i am doing a bitwise AND operation but i am getting wrong data.
elaborating on this:
$value have a hex value 0xffffffc0.
And i have done some grepping and i have hex value stored in $pattern.
so i need to do and operation of $value & $pattern but getting wrong data, because whatever the $value i am passing its getting converted into decimal value.
$value = hex '0xffffffc0' ;
print $value;
$pattern = 0x35040 ; ####this value will be from the grepping####
$pattern1 = $pattern & 0xfffffffc;
print "$pattern1\n";
i am seeing O/P: 35008
You can specify hexadecimal numbers like $hex_number = 0xaf56, hex converts a string having hexadecimal text into the equivalent number, i.e. $hex_number == hex 'af56'. Numbers are represented in binary inside a computer and they will behave properly with any bit-wise operation you do on them.
It is only when you print a number that you need to be careful about how you want it formatted. By default numbers are printed in decimal, you can print in hexadecimal using the %x format specifier with printf.
Read a file that contains an address and a data, like below:
#0, 12345678
#1, 5a5a5a5a
...
My aim is to read the address and the data. Consider the data I read is in hex format, and then I need to unpack them to binary number.
So 12345678 would become 00010010001101000101011001111000
Then, I need to further unpack the transferred binary number to another level.
So it becomes, 00000000000000010000000000010000000000000001000100000001000000000000000100000001000000010001000000000001000100010001000000000000
They way I did is like below
while(<STDIN>) {
if (/\#(\S+)\s+(\S+)/) {
$addr = $1;
$data = $2;
$mem{$addr} = ${data};
}
}
foreach $key (sort {$a <=> $b} (keys %mem)) {
my $str = unpack ('B*', pack ('H*',$mem{$key}));
my $str2 = unpack ('B*', pack ('H*', $str));
printf ("#%x ", $key);
printf ("%s",$str2);
printf ("\n");
}
It works, however, my next step is to do some numeric operation on the transferred bits.
Such as bitwise or and shifting. I tried << and | operator, both are for numbers, not strings. So I don't know how to solve this.
Please leave your comments if you have better ideas. Thanks.
You can employ Bit::Vector module from metaCPAN
use strict;
use warnings;
use Bit::Vector;
my $str = "1111000011011001010101000111001100010000001111001010101000111010001011";
printf "orig str: %72s\n", $str;
#only 72 bits for better view
my $vec = Bit::Vector->new_Bin(72,$str);
printf "vec : %72s\n", $vec->to_Bin();
$vec->Move_Left(2);
printf "left 2 : %72s\n", $vec->to_Bin();
$vec->Move_Right(4);
printf "right 4 : %72s\n", $vec->to_Bin();
prints:
orig str: 1111000011011001010101000111001100010000001111001010101000111010001011
vec : 001111000011011001010101000111001100010000001111001010101000111010001011
left 2 : 111100001101100101010100011100110001000000111100101010100011101000101100
right 4 : 000011110000110110010101010001110011000100000011110010101010001110100010
If you need do some math with arbitrary precision, you can also use Math::BigInt or use bigint (http://perldoc.perl.org/bigint.html)
Hex and binary are text representation of numbers. Shifting and bit manipulations are numerical operations. You want a number, not text.
my $hex = '5a5a5a5a';
$num = hex($hex); # Convert to number.
$num >>= 1; # Manipulate the number.
$hex = sprintf('%08X', $num); # Convert back to hex.
In a comment, you mention you want to deal with 256 bit numbers. The native numbers don't support that, but you can use Math::BigInt.
My final solution of this is forget about treat them as numbers, just treat them as string . I use substring and string concentration instead of shift. Then for the or operation , I just add each bit of the string, if it's 0 the result is 0, else is 1.
It may not be the best way to solve this problem. But that's the way I finally used.
I have a problem understanding and using the 'vec' keyword.
I am reading a logpacket in which values are stored in little endian hexadecimal. In my code, I have to unpack the different bytes into scalars using the unpack keyword.
Here's an example of my problem:
my #hexData1 = qw(50 65);
my $data = pack ('C*', #hexData1);
my $x = unpack("H4",$data); # At which point the hexadecimal number became a number
print $x."\n";
#my $foo = sprintf("%x", $foo);
print "$_-> " . vec("\x65\x50", $_, 1) . ", " for (0..15); # This works.
print "\n";
But I want to use the above statement in the way below. I don't want to send a string of hexadecimal in quotes. I want to use the scalar array of hex $x. But it won't work. How do I convert my $x to a hexadecimal string. This is my requirement.
print "$_-> " . vec($x, $_, 1).", " for (0..15); # This doesn't work.
print "\n";
My final objective is to read the third bit from the right of the two byte hexadecimal number.
How do I use the 'vec' command for that?
You are making the mistake of unpacking $data into $x before using it in a call to vec. vec expects a string, so if you supply a number it will be converted to a string before being used. Here's your code
my #hexData1 = qw(50 65);
my $data= pack ('C*', #hexData1);
The C pack format uses each value in the source list as a character code. It is the same as calling chr on each value and concatenating them. Unfortunately your values look like decimal, so you are getting chr(50).chr(65) or "2A". Since your values are little-endian, what you want is chr(0x65).chr(0x50) or "\x65\x50", so you must write
my $data= pack ('(H2)*', reverse #hexData1);
which reverses the list of data (to account for it being little-endian) and packs it as if it was a list of two-digit hex strings (which, fortunately, it is).
Now you have done enough. As I say, vec expects a string so you can write
print join ' ', map vec($data, $_, 1), 0 .. 15;
print "\n";
and it will show you the bits you expect. To extract the the 3rd bit from the right (assuming you mean bit 13, where the last bit is bit 15) you want
print vec $data, 13, 1;
First, get the number the bytes represent.
If you start with "\x50\x65",
my $num = unpack('v', "\x50\x65");
If you start with "5065",
my $num = unpack('v', pack('H*', "5065"));
If you start with "50","65",
my $num = unpack('v', pack('H*', join('', "50","65"));
Then, extract the bit you want.
If you want bit 10,
my $bit = ($num >> 10) & 1;
If you want bit 2,
my $bit = ($num >> 2) & 1;
(I'm listing a few possibilities because it's not clear to me what you want.)