In my script i am dealing with binary value and i need to replace 0->1 and 1->0 at one place.
example :
input digit = 10101001
output digit = 01010110
I tried $string =~ s/1/0/; and reverse function but that is getting fail to give me correct out put.
can some one help me out.
Use tr:
my $str = '10101001';
$s =~ tr/01/10/;
print "$s\n";
Outputs:
01010110
If your input string has only those two possibilities 0 and 1, then you can use substitution in a multi-stage approach:
$str =~ s/1/x/g; # all 1's to x
$str =~ s/0/1/g; # all 0's to 1
$str =~ s/x/0/g; # all x's to 0
This is not a bad option for languages that only provide substitutions, but Perl also has an atomic translation feature:
$str =~ tr/01/10/;
which will work just as well (better, really, since it's less code and probably less passes over the data).
You could also go mathy on this and use the bitwise XOR operator ^...
my $input = '10101001';
my $binval = oct( '0b'.$input );
my $result = $binval ^ 0b11111111;
printf "%08b\n", $result;
...which will also give you 01010110.
This of course has the downside of being dependent on the length of the bit input string. The given solution only works for 8-bit values. It wouldn't be hard to generalize for any number of bits, though.
To incorporate Lưu Vĩnh Phúc's comment - you can also use the bitwise NOT operator ~. Again, the implementation is dependent on the number of bits as you need to truncate the result:
my $input = '10101001';
my $binval = oct( '0b'.$input );
print substr( sprintf ( '%b', ~$binval ), -8 )."\n";
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'm trying to truncate a string in a select input option using perl if it is longer than a set value, though i can't get it to work correctly.
my $value = defined $option->{value} ? $option->{value} : '';
my $maxValueLength = 50;
if ($value.length > $maxValueLength) {
$value = substr $value, 0, $maxValueLength + '...';
}
Another option is regex
$string =~ s/.{$maxLength}\K.*/.../;
It matches any character (.) given number of times ({N}, here $maxLength), what is the first $maxLength characters in $string; then \K makes it "forget" all previous matches so those won't get replaced later. The rest of the string that is matched is then replaced by ...
See Lookaround assertions in perlre for \K.
This does start the regex engine for a simple task but it doesn't need any conditionals -- if the string is shorter than the maximum length the regex won't match and nothing happens.
Your code has several syntax errors. Turn on use strict and use warnings if you don't have it, and then read the error messages it tells you about. This is a bit tricky because of Perl's very complex syntax (see also Damian Conway's keynote from the 2020 Perl and Raku Conference), but it boils down to these:
Use of uninitialized value in concatenation (.) or string at line 7
Argument "..." isn't numeric in addition (+) at line 8
I've used the following adaption of your code to produce these
use strict;
use warnings;
my $value = '1234567890' x 10;
my $maxValueLength = 50;
if ( $value.length > $maxValueLength ) {
$value = substr $value, 0, $maxValueLength + '...';
}
print $value;
Now let's see what they mean.
The . operator in Perl is a concatenation. You cannot use it to call methods, and length is not a method on a string. Perl thinks you are using the built-in length (a function, not a method) without an argument, which makes it default to $_. Most built-ins do this, to make one-liners shorter. But $_ is not defined. Now the . tries to concatenate the length of undef to $value. And using undef in a string operation leads to this warning.
The correct way of doing this is length $value (or with parentheses if you prefer them, length($value)).
The + operator is not concatenation (we just learned that the . is). It's a numerical addition. Perl is pretty good at converting between strings and numbers as there aren't really any types, so saying 1 + "5" would give you 6 without problems, but it cannot do that for a couple of dots in a string. Hence it complains about a non-number value in an addition.
You want the substring with a given length, and then you want to attach the three dots. Because of associativity (or stickyness) of operators you will need to use parentheses () for your substr call.
$value = substr($value, 0, $maxValueLength) . '...';
To find a length of the string use length(STRING)
Here is the code snippet how you can modify the script.
#!/usr/bin/perl
use strict;
use warnings;
use feature qw(say);
my $string = "abcdefghijklmnopqrstuvwxyz abcdefghijklmnopqrstuvwxyz abcdefghijklmnopqrstuvwxyz";
say "length of original string is:".length($string);
my $value = defined $string ? $string : '';
my $maxValueLength = 50;
if (length($value) > $maxValueLength) {
$value = substr $value, 0, $maxValueLength;
say "value:$value";
say "value's length:".length($value);
}
Output:
length of original string is:80
value:abcdefghijklmnopqrstuvwxyz abcdefghijklmnopqrstuvw
value's length:50
I am after some help trying to convert the following log I have to plain text.
This is a URL so there maybe %20 = 'space' and other but the main bit I am trying convert is the char(1,2,3,4,5,6) to text.
Below is an example of what I am trying to convert.
select%20char(45,120,49,45,81,45),char(45,120,50,45,81,45),char(45,120,51,45,81,45)
What I have tried so far is the following while trying to added into the char(in here) to convert with the chr($2)
perl -pe "s/(char())/chr($2)/ge"
All this has manage to do is remove the char but now I am trying to convert the number to text and remove the commas and brackets.
I maybe way off with how I am doing as I am fairly new to to perl.
perl -pe "s/word to remove/word to change it to/ge"
"s/(char(what goes in here))/chr($2)/ge"
Output try to achieve is
select -x1-Q-,-x2-Q-,-x3-Q-
Or
select%20-x1-Q-,-x2-Q-,-x3-Q-
Thanks for any help
There's too much to do here for a reasonable one-liner. Also, a script is easier to adjust later
use warnings;
use strict;
use feature 'say';
use URI::Escape 'uri_unescape';
my $string = q{select%20}
. q{char(45,120,49,45,81,45),char(45,120,50,45,81,45),}
. q{char(45,120,51,45,81,45)};
my $new_string = uri_unescape($string); # convert %20 and such
my #parts = $new_string =~ /(.*?)(char.*)/;
$parts[1] = join ',', map { chr( (/([0-9]+)/)[0] ) } split /,/, $parts[1];
$new_string = join '', #parts;
say $new_string;
this prints
select -x1-Q-,-x2-Q-,-x3-Q-
Comments
Module URI::Escape is used to convert percent-encoded characters, per RFC 3986
It is unspecified whether anything can follow the part with char(...)s, and what that might be. If there can be more after last char(...) adjust the splitting into #parts, or clarify
In the part with char(...)s only the numbers are needed, what regex in map uses
If you are going to use regex you should read up on it. See
perlretut, a tutorial
perlrequick, a quick-start introduction
perlre, the full account of syntax
perlreref, a quick reference (its See Also section is useful on its own)
Alright, this is going to be a messy "one-liner". Assuming your text is in a variable called $text.
$text =~ s{char\( ( (?: (?:\d+,)* \d+ )? ) \)}{
my #arr = split /,/, $1;
my $temp = join('', map { chr($_) } #arr);
$temp =~ s/^|$/"/g;
$temp
}xeg;
The regular expression matches char(, followed by a comma-separated list of sequences of digits, followed by ). We capture the digits in capture group $1. In the substitution, we split $1 on the comma (since chr only works on one character, not a whole list of them). Then we map chr over each number and concatenate the result into a string. The next line simply puts quotation marks at the start and end of the string (presumably you want the output quoted) and then returns the new string.
Input:
select%20char(45,120,49,45,81,45),char(45,120,50,45,81,45),char(45,120,51,45,81,45)
Output:
select%20"-x1-Q-","-x2-Q-","-x3-Q-"
If you want to replace the % escape sequences as well, I suggest doing that in a separate line. Trying to integrate both substitutions into one statement is going to get very hairy.
This will do as you ask. It performs the decoding in two stages: first the URI-encoding is decoded using chr hex $1, and then each char() function is translated to the string corresponding to the character equivalents of its decimal parameters
use strict;
use warnings 'all';
use feature 'say';
my $s = 'select%20char(45,120,49,45,81,45),char(45,120,50,45,81,45),char(45,120,51,45,81,45)';
$s =~ s/%(\d+)/ chr hex $1 /eg;
$s =~ s{ char \s* \( ( [^()]+ ) \) }{ join '', map chr, $1 =~ /\d+/g }xge;
say $s;
output
select -x1-Q-,-x2-Q-,-x3-Q-
I'm trying to find the index of white space in a string in Perl.
For example, if I have the string
stuff/more stuffhere
I'd like to select the word "more" with a substring method. I can find the index of "/" but haven't figured out how to find the index of white space. The length of the substring I'm trying to select will vary, so I can't hard code the index. There will only be one white space in the string (other than those after the end of the string).
Also, if anybody has any better ideas of how to do this, I'd appreciate hearing them. I'm fairly new to programming so I'm open to advice. Thanks.
Just use index:
#!/usr/bin/perl
use warnings;
use strict;
use feature qw{ say };
my $string = 'stuff/more stuffhere';
my $index_of_slash = index $string, '/';
my $index_of_space = index $string, ' ';
say "Between $index_of_slash and $index_of_space.";
The output is
Between 5 and 10.
Which is correct:
0 1
01234567890123456789
stuff/more stuffhere
If by "whitespace" you also mean tabs or whatever, you can use a regular expression match and the special variables #- and #+:
#!/usr/bin/perl
use warnings;
use strict;
use feature qw{ say };
my $string = "stuff/more\tstuffhere";
if ($string =~ m{/.*(?=\s)}) {
say "Between $-[0] and $+[0]";
}
The (?=\s) means is followed by a whitespace character, but the character itself is not part of the match, so you don't need to do any maths on the returned values.
As you stated, you want to select the word between the first /
and the first space following it.
If this is the case, you maybe don't need any index (you need just
the word).
A perfect tool to find something in a text is regex.
Look at the following code:
$txt = 'stuff/more stuffxx here';
if ($txt =~ /\/(.+?) /) {
print "Match: $1.\n";
}
The regex used tries to match:
a slash,
a non-empty sequence of any chars (note ? - reluctant
version), enclosed in a capturing group,
a space.
So after the match $1 contains what was captured by the first
capturing group, i.e. "your" word.
But if for any reason you were interested in starting and ending
offsets to this word, you can read them from $-[1]
and $+[1] (starting / ending indices of the first capturing group).
The arrays #- (#LAST_MATCH_START) and #+ (#LAST_MATCH_END) give offsets of the start and end of last successful submatches. See Regex related variables in perlvar.
You can capture your real target, and then read off the offset right after it with $+[0]
#+
This array holds the offsets of the ends of the last successful submatches in the currently active dynamic scope. $+[0] is the offset into the string of the end of the entire match. This is the same value as what the pos function returns when called on the variable that was matched against.
Example
my $str = 'target and target with spaces';
while ($str =~ /(target)\s/g)
{
say "Position after match: $+[0]"
}
prints
Position after match: 7
Position after match: 18
These are positions right after 'target', so of spaces that come after it.
Or you can capture \s instead and use $-[1] + 1 (first position of the match, the space).
You can use
my $str = "stuff/more stuffhere";
if ($str =~ m{/\K\S+}) {
... substr($str, $-[0], $+[0] - $-[0]) ...
}
But why substr? That's very weird there. Maybe if you told us what you actually wanted to do, we could provide a better alternatives. Here are three cases:
Data extraction:
my $str = "stuff/more stuffhere";
if ( my ($word) = $str =~ m{/(\S+)} ) {
say $word; # more
}
Data replacement:
my $str = "stuff/more stuffhere";
$str =~ s{/\K\S+}{REPLACED};
say $str; # stuff/REPLACED stuffhere
Data replacement (dynamic):
my $str = "stuff/more stuffhere";
$str =~ s{/\K(\S+)}{ uc($1) }e;
say $str; # stuff/MORE stuffhere
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.