#!/usr/bin/perl

use strict;
use warnings;

package aaa; {
    use overload (
        '""'   => sub { return(1000000000000000) },
        '0+'   => sub { return(1000000000000000) },
        'fallback' => 1
    );

    sub new { return (bless({}, shift)); }
}

package bbb; {
    use overload (
        '""'   => sub { return(999999999999999) },
        '0+'   => sub { return(999999999999999) },
        'fallback' => 1
    );

    sub new { return (bless({}, shift)); }
}

package main; {
    print("Object overloaded as 16-digit integer\n");
    my $a = aaa->new();
    print("STRING: $a\n");
    print('NUMBER: ', 0 + $a, "\n\n");

    print("Object overloaded as 15-digit integer\n");
    my $b = bbb->new();
    print("STRING: $b\n");
    print('NUMBER: ', 0 + $b, "\n\n");

    print("Addition with 16-digit integer\n");
    print('NUMBER: ', 0 + 1000000000000000, "\n");
}
[download]

Object overloaded as 16-digit integer
STRING: 1000000000000000
NUMBER: 1e+15

Object overloaded as 15-digit integer
STRING: 999999999999999
NUMBER: 999999999999999

Addition with 16-digit integer
NUMBER: 1000000000000000

Update:Using int(0 + $a) does restore the result back to an integer, but that doesn't address my desire to negate the floating-point conversion in the first place.