HaloO,
Mark J. Reed wrote:
For any numeric type of $x, $x++ should mean $x += 1.3.14 becomes
4.14. -3.14 becomes -2.14 (which indicates that floor() is not
involved) . 5/8 becomes 13/8. The step size is irrelevant. If $x is
so large that adding 1 gets lost due to the precision, then OK, ++
On Thu, Jul 10, 2008 at 12:45 PM, Jon Lang <[EMAIL PROTECTED]> wrote:
> So if '++' works on strings, it might be reasonable to have "bob"++ == "boc".
I was assuming that would continue to be true, as it is currently in
Perl5, Pugs, and Rakudo.
(Well, technically "bob"++ is an error, but { my $x =
Mark J. Reed wrote:
> All of this is IMESHO, of course, but I feel rather strongly on this
> issue. " ++ " means " += 1 ".
Agreed. Anything else violates the principle of least surprise.
Mind you, this is only true for numerics, where the concept of "1"
potentially has meaning. For non-numerics
2008/7/10 TSa <[EMAIL PROTECTED]>:
> So, does it make sense to define ++ and -- for nums as
> $x++ meaning $x += $x.step?
No.
> Or should these operators floor the value to an Int and step the result?
No.
For any numeric type of $x, $x++ should mean $x += 1.3.14 becomes
4.14. -3.14 becomes
2008/7/10 TSa <[EMAIL PROTECTED]>:
> HaloO,
>
> Larry Wall wrote:
>> Well, maybe 0 .. 10-ε or some such.
>
> This ε there is what I have as the .step method of nums
> in the thread "The Inf type". That is $min..^$max is the
> same as $min..($max-$max.step). For Ints the .step is
> always 1. For Num
HaloO,
Larry Wall wrote:
> Well, maybe 0 .. 10-ε or some such.
This ε there is what I have as the .step method of nums
in the thread "The Inf type". That is $min..^$max is the
same as $min..($max-$max.step). For Ints the .step is
always 1. For Nums it depends on the number, that is the
spacing of
