Re: Rounding with Div and Mod operators

>[This has been debated at length over the years, and I've found the
>argument to be similar to the byte order argument in that there are
>advantages and disadvantages to both, but none strong enough to tip
>the scales firmly in favor of one or the other. For what it's worth,
>Fortran 90 uses round toward zero. -John]


Ada has both a mod operator and a rem operator, probably because the
designers thought hard about this. Quoting from the language reference
manual <http://www.adaic.org/standards/83lrm/html/lrm-04-05.html#4.5.5>:+
>>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. Integer division satisfies the identity


        (-A)/B = -(A/B) = A/(-B)


The result of the modulus operation is such that (A mod B) has the sign of
B and an absolute value less than the absolute value of B; in addition,
for
some integer value N, this result must satisfy the relation


        A = B*N + (A mod B)


<<


I guess a search in the literature around the development of Ada will
reveal some of the discussions leading to the inclusion of both mod and
rem.


-Lasse

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