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.)
Related
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.
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'm trying to write a script to find hex strings in a text file and convert them to their reverse byte order. The trouble I'm having is that some of the hex strings are 16 bit and some are 64 bits. I've used Perl's pack to pack and unpack the 16 bit hex numbers and that works fine, but the 64 bit does not.
print unpack("H*", (pack('I!', 0x20202032))). "\n"; #This works, gives 32202020
#This does not
print unpack("H*", (pack('I!', 0x4f423230313430343239303030636334))). "\n";
I've tried the second with the q and Q (where I get ffffffffffffffff). Am I approaching this all wrong?
As bit of background, I've got a multi-gigabyte pipe-delimited text file that has hex strings in reverse byte order as explained above. Also, the columns of the file are not standard; sometimes the hex strings appear in one column, and sometimes in another. I need to convert the hex strings to their reverse byte order.
Always use warnings;. If you do, you'll get the following message:
Integer overflow in hexadecimal number at scratch.pl line 8.
Hexadecimal number > 0xffffffff non-portable at scratch.pl line 8.
These can be resolved by use bigint; and by changing your second number declaration to hex('0x4f423230313430343239303030636334').
However, that number is still too large for pack 'I' to be able to handle.
Perhaps this can be done using simple string manipulation:
use strict;
use warnings;
my #nums = qw(
0x20202032
0x4f423230313430343239303030636334
);
for (#nums) {
my $rev = join '', reverse m/([[:xdigit:]]{2})/g;
print "$_ -> 0x$rev\n"
}
__END__
Outputs:
0x20202032 -> 0x32202020
0x4f423230313430343239303030636334 -> 0x3463633030303932343034313032424f
Or to handle digits of non-even length:
my $rev = $_;
$rev =~ s{0x\K([[:xdigit:]]*)}{
my $hex = $1;
$hex = "0$hex" if length($hex) % 2;
join '', reverse $hex =~ m/(..)/g;
}e;
print "$_ -> $rev\n"
To be pedantic, the hex numbers in your example are 32-bit and 128-bit long, not 16 and 64. If the longest one was only 64-bit long, you could successfully use the Q pack template as you supposed (provided hat your perl has been compiled to support 64-bit integers).
The pack/unpack solution can be used anyway (if with the addition of a reverse - you also have to remove the leading 0x from the hex strings or trim the last two characters from the results):
print unpack "H*", reverse pack "H*", $hex_string;
Example with your values:
perl -le 'print unpack "H*", reverse pack "H*", "4f423230313430343239303030636334"'
3463633030303932343034313032424f
From this question:
bytearray - Perl pack/unpack and length of binary string - Stack Overflow
I've learned that #unparray = unpack("d "x5, $aa); in the snippet below results with string items in the unparray - not with double precision numbers (as I expected).
Is it possible to somehow obtain an array of double-precision values from the $aa bytestring in the snippet below?:
$a = pack("d",255);
print length($a)."\n";
# prints 8
$aa = pack("ddddd", 255,123,0,45,123);
print length($aa)."\n";
# prints 40
#unparray = unpack("d "x5, $aa);
print scalar(#unparray)."\n";
# prints 5
print length($unparray[0])."\n"
# prints 3
printf "%d\n", $unparray[0] '
# prints 255
# one liner:
# perl -e '$a = pack("d",255); print length($a)."\n"; $aa = pack("ddddd", 255,123,0,45,123); print length($aa)."\n"; #unparray = unpack("d "x5, $aa); print scalar(#unparray)."\n"; print length($unparray[0])."\n" '
Many thanks in advance for any answers,
Cheers!
What makes you think it's not stored as a double?
use feature qw( say );
use Config qw( %Config );
use Devel::Peek qw( Dump );
my #a = unpack "d5", pack "d5", 255,123,0,45,123;
say 0+#a; # 5
Dump $a[0]; # NOK (floating point format)
say $Config{nvsize}; # 8 byte floats on this build
Sorry, but you've misunderstood hobbs' answer to your earlier question.
$unparray[0] is a double-precision floating-point value; but length is not like (say) C's sizeof operator, and doesn't tell you the size of its argument. Rather, it converts its argument to a string, and then tells you the length of that string.
For example, this:
my $a = 3.0 / 1.5;
print length($a), "\n";
will print this:
1
because it sets $a to 2.0, which gets stringified as 2, which has length 1.
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)