I need to convert a number from decimal to binary in Perl where my constraint is that the binary number width is set by a variable:
for (my $i = 0; $i<32; $i++)
{
sprintf("%b",$i) # This will give me a binary number whose width is not fixed
sprintf("%5b",$i) # This will give me binary number of width 5
# Here is what I need:
sprintf (%b"$MY_GENERIC_WIDTH"b, $i)
}
I can probably use a work-around in my print statements, but the code would be much cleaner if I can do the aforementioned.
Your question amounts to the following:
How do I build the string %5b where 5 is variable?
Using concatenation.
"%".$width."b"
That can also be written as
"%${width}b"
In more complex cases, you might want to use the following, but it's overkill here.
join('', "%", $width, "b")
Note that sprintf accepts a * as a placeholder for a value to be provided in a variable.
sprintf("%*b", $width, $num)
If you want leading zeroes instead of leading spaces, just add a 0 immediately after the %.
You can interpolate the width into the format string:
my $width = 5;
for my $i (0..31) {
printf "%${width}b\n", $i;
}
Or use a * to input it via a variable:
my $width = 5;
for my $i (0..31) {
printf "%*b\n", $width, $i;
}
Both outputs:
0
1
10
11
100
101
110
111
1000
1001
1010
1011
1100
1101
1110
1111
10000
10001
10010
10011
10100
10101
10110
10111
11000
11001
11010
11011
11100
11101
11110
11111
Related
when I try to execute the following(01434.210 instead of 1434.210)
$val=22749.220-(21315.010+01434.210)
print $val
I get these output
output 638.207900000001
But according to me output must be 0.
What am I missing?
A leading 0 in a literal number makes Perl interpret the value I'm base 8:
123 # 123, in decimal
0123 # 123 in octal, but 83 in decimal
This isn't the same for strings converted to numbers. In those Perl ignores the leading 0s. The string-to-number conversion only deals in base-10:
"123" + 0 # 123
"0123" + 0 # still 123
In your example in the comment, you convert a literal number to a string with a leading zero. When you convert that string back to its numeric form you get the same value you started with:
$val=sprintf("%05d",1434); # converting 1434 to the string "01434"
print $val; print "\n"; # still a string
print $val+21315; # "01434" + 21315 => 1434 + 21315
print "\n";
print 01434+21315; # oct(1434) + 21315 => 796 + 21315
The leading zero notation helps with certain builtins that typically use octal numbers, such as those that deal with unix permissions:
chmod 0644, #files
Input log file:
Nservdrx_cycle 4 servdrx4_cycle
HCS_cellinfo_st[10] (type = (LTE { 2}),cell_param_id = (28)
freq_info = (10560),band_ind = (rsrp_rsrq{ -1}),Qoffset1 = (0)
Pcompensation = (0),Qrxlevmin = (-20),cell_id = (7),
agcreserved{3} = ({ 0, 0, 0 }))
channelisation_code1 16/5 { 4} channelisation_code1
sync_ul_info_st_ (availiable_sync_ul_code = (15),uppch_desired_power =
(20),power_ramping_step = (3),max_sync_ul_trans = (8),uppch_position_info =
(0))
trch_type PCH { 7} trch_type8
last_report 0 zeroth bit
I was trying to extract only integer for my above inputs but I am facing some
issue with if the string contain integer at the beginning and at the end
For ( e.g agcreserved{3},HCS_cellinfo_st[10],Qoffset1)
here I don't want to ignore {3},[10] and 1 but in my code it does.
since I was extracting only integer.
Here I have written simple regex for extracting only integer.
MY SIMPLE CODE:
use strict;
use warnings;
my $Ipfile = 'data.txt';
open my $FILE, "<", $Ipfile or die "Couldn't open input file: $!";
my #array;
while(<$FILE>)
{
while ($_ =~ m/( [+-]?\d+ )/xg)
{
push #array, ($1);
}
}
print "#array \n";
output what I am getting for above inputs:
4 4 10 2 28 10560 -1 1 0 0 -20 7 3 0 0 0 1 16 5 4 1 15 20 3 8 0 7 8 0
expected output:
4 2 28 10560 -1 0 0 -20 7 0 0 0 4 15 20 3 8 0 7 0
If some body can help me with explanation ?
You are catching every integer because your regex has no restrictions on which characters can (or can not) come before/after the integer. Remember that the /x modifier only serves to allow whitespace/comments inside your pattern for readability.
Without knowing a bit more about the possible structure of your output data, this modification achieves the desired output:
while ( $_ =~ m! [^[{/\w] ( [+-]?\d+ ) [^/\w]!xg ) {
push #array, ($1);
}
I have added rules before and after the integer to exclude certain characters. So now, we will only capture if:
There is no [, {, /, or word character immediately before the number
There is no / or word character immediately after the number
If your data could have 2-digit numbers in the { N} blocks (e.g. PCH {12}) then this will not capture those and the pattern will need to become much more complex. This solution is therefore quite brittle, without knowing more of the rules about your target data.
I have to write a Perl script that converts a binary number, specified as an
argument, to a decimal number. In the question there's a hint to use the reverse function.
We have to assume that the binary number is in this format
EDIT: This is what I've progressed to (note this is code from my textbook that I've messed with):
#!/usr/bin/perl
# dec2.pl: Converts decimal number to binary
#
die("No arguments\n") if ( $#ARGV == -1 ) ;
foreach $number (#ARGV) {
$original_number = $number ;
until ($number == 0 ) {
$bit = $number % 2 ;
unshift (#bit_arr, $bit) ;
$number = int($number / 2 );
}
$binary_number = join ("", #bit_arr) ;
print reverse ("The decimal number of $binary_number is $original_number\n");
$#bit_arr = -1;
}
When executed:
>./binary.pl 8
The decimal number of 1000 is 8
I don't know how to word it to make the program know to add up all of the 1's in the number that is inputted.
You could just use sprintf to do the converting for you...
sprintf("%d", 0b010101); # Binary string 010101 -> Decimal 21
sprintf("%b", 21) # Decimal 21 -> Binary 010101 string
Of course, you can also just eval a binary string with 0b in front to indicate binary:
my $binary_string = '010101';
my $decimal = eval("0b$binary"); # 21
You don't have to use reverse, but it makes it easy to think about the problem with respect to exponents and array indices.
use strict;
use warnings;
my $str = '111110100';
my #bits = reverse(split(//, $str));
my $sum = 0;
for my $i (0 .. $#bits) {
next unless $bits[$i];
$sum += 2 ** $i;
}
First of all, you are suppose to convert from a binary to decimal, not the other way around, which you means you take an input like $binary = '1011001';.
The first thing you need to do is obtain the individual bits (a0, a1, etc) from that. We're talking about splitting the string into its individual digits.
for my $bit (split(//, $binary)) {
...
}
That should be a great starting point. With that, you have all that you need to apply the following refactoring of the formula you posted:
n = ( ( ( ... )*2 + a2 )*2 + a1 )*2 + a0
[I have no idea why reverse would be recommended. It's possible to use it, but it's suboptimal.]
For example:
I have $a= -1. If I print it using printf with %.4b or %b, it gives me 32-bit all 1's.
But, I only want to print the least significant 4 bits like 1111 in the file in binary.
Any ideas how to do it?
Thanks
-1 in binary is represented via 2s complement, so it is all 1s. (See here for more: What is “2's Complement”?)
If you want to 'limit' it, then the way you can do this is with a bitwise and.
Switching on 4 bits is
1+2+4+8 = 15.
Therefore:
use strict;
use warnings;
my $val = -1;
printf ( "%b", $val & 15 );
%.4b refers to fractional digits, %04b formats to at least 4 digits, padding leading 0s as needed.
To cater for negative integers, take the modulus by 16 ( 2^<number of least significant bits> ).
my #b = (12, 59, -1, 1 ); # sample of integers
#b = map { $_ % 16; } #b; # take modulus
printf ("4-bits: %04b" . (", %04b" x $#b) . ";\n", #b );
# output with computed number of placeholders
Consider this short example:
$a = pack("d",255);
print length($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", $unparray[0] '
# Prints 255
# As a one-liner:
# perl -e '$a = pack("d",255); print length($a)."\n"; $aa = pack("dd", 255,123,0,45,123); print length($aa)."\n"; #unparray = unpack("d "x5, $aa); print scalar(#unparray)."\n"; print length($unparray[0])."\n"; printf "%d\n", $unparray[0] '
Now, I'd expect a double-precision float to be eight bytes, so the first length($a) is correct. But why is the length after the unpack (length($unparray[0])) reporting 3 - when I'm trying to go back the exact same way (double-precision, i.e. eight bytes) - and the value of the item (255) is correctly preserved?
By unpacking what you packed, you've gotten back the original values, and the first value is 255. The stringification of 255 is "255", which is 3 characters long, and that's what length tells you.