HaloO,
Smylers wrote:
BTW, could we define that the arithmetic shift ops do just that,
whereas the string ones do logical shift? And in addition that for
the bit inversion +^$a == -1 - $a holds? Note that -1 == +^0.
Does that assume a two's complement system? Is that a safe assumption
to make about everywhere Perl 6 will run? (Is it even a safe assumption
to make about Perl 5?)
Well, there are two ways of conduct. Either the spec prescribes the
behavior on the language level or it leaves it undefined. The former
should go with the equalities I gave, the latter is a bad idea for a
high level language. I mean letting platform details shine through
only makes it convenient for the implementor but harder for users.
Note further that in infinite precision the arithmetic shift left
maintains the sign ...
Do we expect Perl 6 to be running on infinite-precision systems?
Well, IIRC the Int type is conceptually infinite. IOW,
my Int $i = 0;
while ++$i { say $i}
should die because of memory exhaustion not end up in a tight
endless loop that wraps around at the max integer. So arithmetic
shift left should preserve sign.
There are other arithmetic properties that I would expect from
a high level language: absence of zero divisors, proper modulo
and integer division including remainder, etc.
Regards,
--
