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Rounding off decimals to the nearest number in PERL

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Top Forums Shell Programming and Scripting Rounding off decimals to the nearest number in PERL
# 8  
Old 03-29-2009
Code:
#!/usr/bin/perl
use POSIX;

my ($nSlices, $bdec, $adec, $rnd);
print "Enter a number = ";
$nSlices1 = <STDIN>;
($bdec, $adec ) = split( /\./, $nSlices1 ); 
$adec = substr($adec,0,1);

print "The value of numbers: before decimal=$bdec after decimal=$adec ";
$rnd = ($adec >= 5) ? ceil($nSlices1) : floor($nSlices1);
print "Rounded to: $rnd\n";

Output:
Code:
$ fc.pl
Enter a number = 5.467
The value of numbers: before decimal=5 after decimal=4 Rounded to: 5
$ fc.pl
Enter a number = 5.647
The value of numbers: before decimal=5 after decimal=6 Rounded to: 6

# 9  
Old 03-30-2009
Thanks rikxik! It works like a treat! Smilie
# 10  
Old 03-30-2009
Quote:
Originally Posted by rdlover
Below is the snippet of the code that i have written. The damn sprintf is not working. No matter what number i enter, it is entering ceil and not floor.Any reasons why ?



Code:
#!/usr/bin/perl

# SOME VARIABLE

use POSIX;

print "Enter a number = ";

$nSlices1 = <STDIN>;

($bdec, $adec ) = split( /\./, $nSlices1 ); # Split the variable at the decimal point and put it into 2 temporary variables


print " The value of numbers before decimal point is $bdec \n";

sprintf("%.3f", $adec) ;

print " The value of numbers after  decimal point is $adec \n";


if ($adec >= 50)

{
     $ceil = ceil($nSlices1); 
     print "The value of $nSlices1 has been rounded off to $ceil \n";
    } 

else 

{
      $floor = floor($nSlices1);
        
       print "The value of input has been rounded off to $floor  \n";
        
}

Code:
#!/usr/bin/perl

# SOME VARIABLE

print "Enter a number = ";

$nSlices1 = <STDIN>;
chomp $nSlices1;
print " The value of numbers before decimal point is $nSlices1 \n";

$nSlices1 = sprintf("%d", $nSlices1);

print " The value of numbers after  decimal point is $nSlices1 \n";
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