JeeBee wrote:
> Dear Perl experts,
Hello,
> The following subroutine contains a while loop that continues to
> multiply $p by 2, until some condition is met.
> (It's for computing a Collatz tree)
> $p turns 0 in this loop, however, and it remains zero for ever, of course.
>
> My question is, what is exactly its maximum value?
$ perl -le'use POSIX;
print for SHRT_MAX, INT_MAX, LONG_MAX, FLT_MAX, DBL_MAX;
'
32767
2147483647
2147483647
3.40282346638529e+38
1.79769313486232e+308
> As I have been browsing
> the web for a while, but am unable to find this information for Perl data
> types. (I know I could perhaps use BigIntegers, but don't know whether
> that would be really necessary).
>
> Is there a difference between 'long' and 'int' or something in Perl???
perldoc perlnumber
> sub left_child_of($) {
> my $p = shift;
> print "Left of $p\n";
> die("3 |/| p-1 (p = $p)") if ($p-1) % 3 != 0;
> $p = ($p-1) / 3;
> print "L$p\n";
> # keep multiplying by 2, until 3|p-1 and 2 |/| p-1
> while((($p-1) % 3 != 0) || (($p-1) % 2 == 0)) {
> print "L$p: 3 |/| p-1\n" if ($p-1) % 3 != 0;
> print "L$p: 2 | p-1\n" if ($p-1) % 2 == 0;
> $p <<= 1;
You are not multiplying $p by 2 like you said which would do what you want:
$p *= 2;
> }
> return $p;
> }
John
--
use Perl;
program
fulfillment
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