My attempt uses the fact that "gb" means "1024 ** 3", whereas "g" means "1000 ** 3", so there is just a diffrence in the bases, whereas the exponents are the same. I just store the single-letter exponent identifiers in a hash as keys with their corresponding exponents as values and do the same with the base identifiers ("" for decimal and "b" for binary) and then do the rest with a regex: 
#!perl -ws
my %exponents = ( 'k' => 1, 'm' => 2, 'g' => 3, 't' => 4 );
my %bases     = ( '' => 1000, 'b' => 1024 );
while (<DATA>) {
   chomp;
   my $old = $_;
   s<(\d+)([a-zA-Z])([a-zA-Z]?)\b>
    < (defined $exponents{lc($2)} and defined $bases{lc($3)}) ? # suff
+ix recognized ?
         $1 * ($bases{lc($3)} ** $exponents{lc($2)}) # substitute if y
+es
      :  "$1$2$3"                                    # otherwise just 
+leave it
    >eg;
   print "`$old' became `$_'\n";
}
__DATA__
My dog makes $30k a year.
I own a 20gb hard drive.
The budget deficite is $10g.
18h is 24d.

[download]
The output is as expected (the last data line does not change).

In reply to Re: converting "30k" string to integer by CombatSquirrel
in thread converting "30k" string to integer by dwhite20899

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