my $calc = ($X-$Y)/$Z;
my $phase = $calc - int $calc;
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Be careful with this technique as storage of decimals in a computer is never accurate and a conversion to int could throw you off.

See this node for a discussion on the same lines

Dividing and format

Specifically the links in my reply here:

Ninthwave's reply in Dividing and format



"No matter where you go, there you are." BB

Be also careful if you have negative numbers...

e.g. $X = 1, $Y = 7, $Z = 5

Then:

($X-$Y) = -6, so $calc = ($X-$Y)/$Z = -1.2

but $phase = $calc - int $calc = -1.2 -1 = -2.2, which is not what you wanted...

--------------------------------
An idea is not responsible for the people who believe in it...

Did you try it?

$ perl -e'my ($X,$Y,$Z)=(1,7,5);my $c=($X-$Y)/$Z;print $c-int$c, $/'
-0.2
$
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That may not be what's wanted, either.

After Compline,
Zaxo

Many Thanks :-)

#! perl -slw
use strict;
use POSIX qw[ modf ];

my $float = 2.543;

my( $fract, $integral ) = modf( $float );
print "$integral - $fract";
__END__
P:\test>462825
2 - 0.543
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What have you tried already?

While Perl is a programming language with many special features, it still has the feature set that you can expect from other programming languages. Even if you don't know how to express this in Perl syntax, you should still be able to come up with a set of instructions that does what you want.

For example, the following set of instructions does what you want and should be easily implemented in Perl:

1. If the absolute of the number x is smaller than 1.0, return the number
2. If the number x is larger than 1, return the phase of the number x-1
3. If the number x is smaller than -1, return the phase of the number x+1

It's easy to convince yourself that this will always produce a solution.

Given that you're dealing with phases, presumably with period 1, if your calculation can produce negative numbers, then I would recommend:

$x - POSIX::floor( $x );
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$x - int( $x );
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( POSIX::modf( $x ) )[ 0 ];
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my $phase;
my $number = 2.543;

#substr
$phase = "0" . substr($number,index($number,"."));

#regex
$number =~ /(\..+)/;
$phase = "0$1";


#int
$phase = $number-int($number);
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If the number doesn't contain a decimal part, you could end up using $1 from a previous match. I'd suggest (maybe some parenthesis can be avoided, but I'm not able to test ATM):

$phase = '0' . (($number =~ /(\..+)/) ? $1 : '');
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I found particularly wit to include the "." inside $1, in order to avoid repetition.

Flavio (perl -e 'print(scalar(reverse("\nti.xittelop\@oivalf")))')

Don't fool yourself.

I can't agree with your solution. The questioner is calculating a phase. Doing math. Your solution is based on string manipulation. Conflating text values and numeric values leads to errors. Refrain from smearing your mental map of a problem space.

Be Appropriate && Follow Your Curiosity

The standard Perl approach to rounding numbers (with sprintf) amounts to using a string manipulation for what is arguably an arithmetic task. Did you have any particular error in mind?

When I first saw the question, the two solutions I thought of were $x - int( $x ) or indeed something along the lines of $x =~ s/\d+(?=\.)//, and I'm curious about what errors could have resulted from the latter approach.

the lowliest monk