Just a nit, for the record, with no great perl relevance:
TSa (Thomas Sandla) 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 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 $y).
Hmm, since 8 - (-3) == 11 this
Mark Reed wrote:
I would really like to see ($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 = $x
[EMAIL PROTECTED] wrote:
There are actuall two usefull definition for %. The first which Ada calls 'mod'
always returns a value 0=XN and yes it has no working value that is an
identity. The other which Ada calls 'rem' defined as follows:
Signed integer division and remainder are defined by
I would really like to see ($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)
