Perhaps bizarrely, tadman's code doesn't work with all numbers. I think it's internal rounding errors on Perl's part rather than a fault with the code (though I could be wrong :). I've broken down the code to analyse what's going on, viz:

$number = 12650 / 10000;
$dp = 2;

print "Actual: $number\n";

my $exp = 10 ** $dp;

$val1 = $number * $exp;
$val2 = $val1 + 0.5;
$val3 = int($val2);
$val4 = $val3 / $exp;

print "val1 = value x exp = $val1\n";
print "val2 = val1 + 0.5 = $val2\n";
print "val3 = int(val2) = $val3\n";
print "val4 = val3 / exp = $val4\n";
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The output:

Actual: 1.265
val1 = value x exp = 126.5
val2 = val1 + 0.5 = 127
val3 = int(val2) = 126
val4 = val3 / exp = 1.26
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Verified on Win32 with ActivePerl 5.8.1 b807, and in Linux with Perl 5.8.0.

Replace the first line with "$number = 1.265;", however, and you get the correct result. Weird.

It's probably storing "127" as 126.9999999999999 which rounds down with the int function. Aren't floating point numbers fun?

The documentation for int suggests that the POSIX::floor function is a better selection for this kind of work.

Ok Goron I brought the int function up sooner to get rid of the floating point asp. And to remove decimels from the addition portion. Seen in this code based of yours.

Code

use strict;

my $number = 12650 / 10000;
#my $number = 1.265;
my $dp = 2;

print "Actual: $number\n";

my $exp = 10 ** ($dp+1);

my $val1 = int($number * $exp);
my $val2 = $val1 + 5;
my $val4 = $val2 / $exp;

print "val1 = value x exp = $val1\n";
print "val2 = val1 + 5 = $val2\n";
#print "val3 = int(val2) = $val3\n";
print "val4 = val2 / exp = $val4\n";
[download]

Output

Actual: 1.265
val1 = value x exp = 1265
val2 = val1 + 5 = 1270
val4 = va2 / exp = 1.27
[download]

Ok I messed it up if you use 1200 you get 12.05 so I think you need to do it in steps throwing away un needed decimals. Oh well need to catch a train will play with it tonight because it is not Internet Explorer and I need something that is not Internet Explorer.

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

The reason this doesn't work is that you're doing the "+5" after the int statement. The purpose of the +5 (or +0.5) is to ensure that the int rounds down anything below x.5 and rounds up anything x.5 and above.

It's so that:

int(0.4 + 0.5) = int(0.9) and therefore 0.4 rounds down to 0.

int(0.5 + 0.5) = int(1.0) and therefore 0.5 rounds up to 1.

How I long for the good old days of MC68000 programming, where there was no such thing as floating point :)

It would seem that, given he wishes not to use printf or formats, his best bet is the sprintf tool.

Eventually you're going to have to use something printf-like function such as sprintf anyway. I was just proposing something that kept the numbers as numbers instead of stringifying them. It is numerically 12.0, even if print decides to show it as 12 without the decimal.

That being said, any time you're dealing with floating point numbers it's pretty much advised to use some kind of formatting method. Some numbers might be assigned a value of 12.0 but end up displaying as 11.9999999999999999981 instead, or even 12.00000000000000012.