Just a nit, for the record, with no great perl relevance:
"TSa (Thomas Sandlaß)" wrote:
But what is the first quarter of year 0? 0.25?
Sure (of course if there were a year 0 instead of becoming 1 BCE)
> And the last quarter of year -1? -0.25?
Sure
> That works numerically, but March of a
year
Mark Reed wrote:
At least, not in cases where the intended result is consistent across 0.
Lots of date arithmetic falls into this category, and works beautifully with
the definitions above.
Does it? If you have a year 0, what is the number corresponding
to the middle of that year? Is it 0.5? Is
On 2005-05-30 05:15, "TSa (Thomas Sandlaß)" <[EMAIL PROTECTED]>
wrote:
> Mark Reed wrote:
>> I would really like to see ($x div $y) be (floor($x/$y))
>
> That is: floor( 8 / (-3) ) == floor( -2. ) == -3
> Or do you want -2?
>
>> and ($x mod $y) be ($x - $x div $y).
>
> Hmm, since 8 - (
Mark Reed wrote:
I would really like to see ($x div $y) be (floor($x/$y))
That is: floor( 8 / (-3) ) == floor( -2. ) == -3
Or do you want -2?
and ($x mod $y) be ($x - $x div $y).
Hmm, since 8 - (-3) == 11 this definition hardly works.
But even with $q = floor( $x / $y ) and $r = $x
I would really like to see ($x div $y) be (floor($x/$y)) and ($x mod $y) be
($x - $x div $y). If the divisor is positive the modulus should be
positive, no matter what the sign of the dividend. Avoids lots of special
case code across 0 boundaries.
On 2005-05-23 18:49, "TSa (Thomas Sandlaß)" <[
