Re^2: long integer to hexadecimal conversion

[kikuchiyo]

I wanted to say the same thing, but I'm late, I guess:

use bigint;

my $n = 123456789123456;
my ($h, $d);
while ($n > 0) {
    $d = $n % 16;
    $n /= 16;
    $h .= sprintf "%x",$d;
}
print scalar reverse $h;

[Joost]

Much simpler:

use bigint; 
my $n = 123456789123456; 
print $n->as_hex;

or just

use bigint;
print 123456789123456->as_hex

[kikuchiyo]

True.

You'd only need my algorithm if you wanted to express the integer in a base other than 16 (or 8).



use bigint; 

my @digits=map chr,(0x30..0x39,0x41..0x5d);
my $base = 22;
my $n = 123456789123456;
my $h;
while($n > 0){
    $h .= $digits[ $n % $base ];
    $n /= $base;
}
print scalar reverse $h';

Result is 4E5G93GHBE8.

[perl_fan]

Super Duper Cool Thanks Joost

[perl_fan]

couldn't have been easier than this. Thanks kikuchiyo, you saved my day