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Postings from May 2005
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)" <Thomas.Sandlass@orthogon.com> wrote: > mark.a.biggar@comcast.net wrote: >> > There are actuall two usefull definition for %. The first which Ada calls >> 'mod' always returns a value 0<=X<N 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 the relation: >> > >> > A = (A/B)*B + (A rem B) >> > >> > where (A rem B) has the sign of A and an absolute value less than the >> absolute value of B. Signed integer division satisfies the identity: > >> > >> > (-A)/B = -(A/B) = A/(-B) >> > >> > It does have a right side identity of +INF. > > This is the truncating div-dominant definition of modulo. > The eulerian definition is mod-dominant and nicely handles > non-integer values. E.g. > > 3.2 == 1.5 * 2 + 0.2 -+--> 3.2 / 1.5 == 2 + 0.2 / 1.5 == 2 + 1/15 > | == 2 + 0.1333... > +--> 3.2 % 1.5 == 0.2 > > Note that -3.2 == -4 + 0.8 == -4.5 + 1.3 == ... > > -3.2 / 1.5 == -3 + 1.3 / 1.5 == -3 + 0.8666... == -2.1333 > -3.2 % 1.5 == 1.3 > > With integers: > > 8 / 3 == (2 + 2/3) == 2 > 8 / (-3) == -(2 + 2/3) == -2 > (-8) / 3 == -(3 - 1/3) == -3 # this might surprise some people ;) > (-8) / (-3) == (3 - 1/3) == 3 > > 8 % (-3) == 8 % 3 == 2 > (-8) % (-3) == (-8) % 3 == 1 # this as well, but it's just -3 * 3 + 1 > > Real valued division can be considered as % 0, that is infinite precision. > While integer arithmetic is % 1. I.e. int $x == $x - $x % 1. > > floor $x == $x - $x % 1 # -1.2 - (-1.2) % 1 == -1.2 - 0.8 == -2 > ceil $x == 1 + floor $x > round $x == floor( $x + 0.5 ) > trunc $x == $x < 0 ?? ceil $x :: floor $x > > To @Larry: how are mod and div defined in Perl6?Thread Previous | Thread Next