Re: Round up 3 values to the next 100

[bobf]

in reply to Round up 3 values to the next 100

See perlfaq4: "Does Perl have a round() function? What about ceil() and floor()? Trig functions?" In particular, take a look at the ceil function in POSIX.

Update: Don't forget about Super Search, which found a particularly relevant thread: Rounding up nearest ten, hundred etc. :-)

[siva kumar]

I hope ceil/floor is used for decimal numbers conversion. But I have only integers that last 3 digits rounded to next 100.

[bobf]

use strict;
use warnings;
use POSIX qw( ceil );

my $round_to = 100;

for( 78345, 5674 )
{
    print "$_ -> ", ceil( $_/$round_to ) * $round_to, "\n";
}

Output:

78345 -> 78400
5674 -> 5700

Take a look at the thread I mentioned in my previous post. There are quite a few examples to help you.

[siva kumar]

Thank you 'bobf'. Worked well ..
Thank you so much.

[siva kumar]

I might have done this

$num = 235634;
 
if (length($num) >=3 ) {
    $var = substr($num,-2,2);
     if ( !$var eq '00')
     {
        $var1 = 100 - $var;
        $num +=  $var1;
     }
     print $num;
}

[bobf]

If you tested that you might be surprised at what it actually does. I'll give you a hint: what does if ( !$var eq '00' ) really mean? (See perlop and look at the table of precedence...)

<Reveal this spoiler or all in this thread>

You might also want to specify how negative numbers should be rounded. If -12345 should round to -12300, my solution will work but the approach you outlined will produce different results.

[gowthamtr]

I guess this will work....
$a is the number to be rounded

#!/usr/bin/perl -w

$a=78345;
$b=100;
$c=$a/$b;
$d=$c;
$c=~s/\..*//;
if( $d > $c ){
$c++;  }
$c*=100;
print $c,"\n";

$c will have the result