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 Nums it depends on the number, that is the
> spacing of Nums is non-uniform but discrete like Ints.
> That is what I call non-uniform Ints in contrast to Rats
> which are dense and their operations ++ and -- are undefined
> unless one wants to jump through the diagonalized Rats
> with skipped duplicates: 1++ -> 1/2, 1/2++ -> 1/3, 1/3++
> -> 2/3, 2/3++ -> 1/4, 1/4++ -> 3/4, etc.
>
> So, does it make sense to define ++ and -- for nums as
> $x++ meaning $x += $x.step? Or should these operators
> floor the value to an Int and step the result?
>
> my $x = 3.14;
> $x++;
> say $x; # num 4.14 or int 4?
>
> $x = 3.14159265358979323846e20;
> $x++;
> say $x; # int 314159265358979323847
> # or int 314159265358979300001
> # or num 3.141592653589794e20 for 64bit float?
>
> The last two options above are because $x.step == 1e5.
> The exact value would need a float with 22 digits mantissa,
> i.e. a quad precision.
>
least surprise (to me) says ++ always increments by 1 for numeric
types. if you want to increment by ε, by all means create ε++ and ε--
ops.

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~jerry