Hello Monks,

More I get closer to my solution more problems I discover. I am trying to convert decimals into 16 bits.

I am using pack and unpack, which works perfect when the first digit of the decimal number is not zero:

I know it does not make sense to have a decimal starting with zero. The reason is that I have a number such as 0.02911 where I am splitting the string before the dot and after so I can send them separately.

So my alternative option would be to convert the decimal to binary B16 send in and then reverse the process. Cookbook 2.4. Converting Between Binary and Decimal

Sample of code with both proposed solutions:

#!/usr/bin/perl
use warnings;
use strict;

my $sampleDecimal = "02911";

my $uint16 = pack 'n', $sampleDecimal;

my $decimal = unpack 'n', $uint16;

print "Decimal: " . $decimal . "\n";

my $binaryStr = desimalToBinary( $sampleDecimal , 16 , 'n' );

print "Binary: " . $binaryStr . "\n";

my $decimalStr = binaryToDecimal( $binaryStr , 16 , 'n' );

print "Decimal from binary: " . $decimalStr . "\n";

sub binaryToDecimal {
    my $binary = shift;
    my $bitSize = shift;
    my $template = shift;
    return unpack($template, pack("B$bitSize", substr("0" x $bitSize .
+ $binary, -$bitSize)));
}

sub desimalToBinary {
    my $decimal = shift;
    my $bitSize = shift;
    my $template = shift;
    my $bitStr = unpack("B$bitSize", pack($template, $decimal));
    $bitStr =~ s/^0+(?=\d)//;   # otherwise you'll get leading zeros
    return $bitStr;
}

__END__

Decimal: 2911
Binary: 101101011111
Decimal from binary: 2911

[download]

Unfortunately, both proposed solutions do give the same result. A non leading zero decimal. I know I could add manually the leading zero, but in cases the number produced with a non zero digit, then I will have completely wrong number.

So the question, is it possible to get the leading zero through pack and unpack using 'n' as a template?

Update 2: Well it seems that the best solution is to use, sprintf as roboticus proposed.

Sample of code/solution:

#!/usr/bin/perl
use warnings;
use strict;

my $sampleDecimal = "02911";

my $uint16 = pack 'n', $sampleDecimal;

my $decimal = unpack 'n', $uint16;

$decimal = sprintf("%05d", $decimal);

print "Decimal: " . $decimal . "\n";

my $binaryStr = desimalToBinary( $sampleDecimal , 16 , 'n' );

print "Binary: " . $binaryStr . "\n";

my $decimalStr = binaryToDecimal( $binaryStr , 16 , 'n' );

$decimalStr = sprintf("%05d", $decimalStr);

print "Decimal from binary: " . $decimalStr . "\n";

sub binaryToDecimal {
    my $binary = shift;
    my $bitSize = shift;
    my $template = shift;
    return unpack($template, pack("B$bitSize", substr("0" x $bitSize .
+ $binary, -$bitSize)));
}

sub desimalToBinary {
    my $decimal = shift;
    my $bitSize = shift;
    my $template = shift;
    my $bitStr = unpack("B$bitSize", pack($template, $decimal));
    $bitStr =~ s/^0+(?=\d)//;   # otherwise you'll get leading zeros
    return $bitStr;
}

__END__

Decimal: 02911
Binary: 101101011111
Decimal from binary: 02911

[download]

Thank you for your time and effort.

Seeking for Perl wisdom...on the process of learning...not there...yet!

In reply to [SOLVED] How to pack and unpack 16 bit decimal with leading zero by thanos1983

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