How do you calculate div and mod of floating point numbers?
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by boost
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Published on 2009-01-30T06:52:21Z
Indexed on
2010/05/09
21:08 UTC
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In Perl, the %
operator seems to assume integers. For instance:
sub foo {
my $n1 = shift;
my $n2 = shift;
print "perl's mod=" . $n1 % $n2, "\n";
my $res = $n1 / $n2;
my $t = int($res);
print "my div=$t", "\n";
$res = $res - $t;
$res = $res * $n2;
print "my mod=" . $res . "\n\n";
}
foo( 3044.952963, 7.1 );
foo( 3044.952963, -7.1 );
foo( -3044.952963, 7.1 );
foo( -3044.952963, -7.1 );
gives
perl's mod=6
my div=428
my mod=6.15296300000033
perl's mod=-1
my div=-428
my mod=6.15296300000033
perl's mod=1
my div=-428
my mod=-6.15296300000033
perl's mod=-6
my div=428
my mod=-6.15296300000033
Now as you can see, I've come up with a "solution" already for calculating div
and mod
. However, what I don't understand is what effect the sign of each argument should have on the result. Wouldn't the div
always be positive, being the number of times n2
fits into n1
? How's the arithmetic supposed to work in this situation?
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