integer division rounding

[Marcin 'Qrczak' Kowalczyk]

18 Jul 2001 16:46:49 +0200, Ralf Muschall <ralf.muschall at alphasat.de> pisze:

> What does this mean?  I always understood C's way of dividing
> negative ints a "do what is the simplest thing for the machine"

It used to be. This is what C++ says:

4 The binary / operator yields the quotient, and the binary  %  operator
  yields  the remainder from the division of the first expression by the
  second.  If the second operand of / or % is zero the behavior is unde-
  fined;  otherwise  (a/b)*b  + a%b is equal to a.  If both operands are
  nonnegative then the remainder is nonnegative; if not, the sign of the
  remainder is implementation-defined70).
  _________________________
  70)  According to work underway toward the revision of ISO C, the pre-
  ferred algorithm for integer division follows the rules defined in the
  ISO  Fortran standard, ISO/IEC 1539:1991, in which the quotient is al-
  ways rounded toward zero.

and C indeed doesn't provide the choice now:

       [#6] When integers are divided, the result of the / operator
       is  the  algebraic  quotient  with   any   fractional   part
       discarded.78)   If  the  quotient  a/b is representable, the
       expression (a/b)*b + a%b shall equal a.
       ____________________

       78)This is often called ``truncation toward zero''.

A consequence of the C definition is that the compiler can't implement
i/2 as i>>1 unless it can be sure that i is nonnegative.

-- 
 __("<  Marcin Kowalczyk * qrczak at knm.org.pl http://qrczak.ids.net.pl/
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