Integer division rounding with negatives in C++
Solution 1:
According to the May 2008 revision,
You're right:
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 undefined; 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-defined75).
Note 75 says:
According to work underway toward the revision of ISO C, the preferred algorithm for integer division follows the rules defined in the ISO Fortran standard, ISO/IEC 1539:1991, in which the quotient is always rounded toward zero.
Chances are that C++ will lag C in this respect. As it stands, it's undefined but they have an eye towards changing it.
I work in the same department as Stroustrup and with a member of the committee. Things take AGES to get accomplished, and its endlessly political. If it seems silly, it probably is.
Solution 2:
As an update to the other answers:
The last draft of C++11, n3242 which is for most practical purposes identical to the actual C++11 standard, says this in 5.6 point 4 (page 118):
For integral operands the / operator yields the algebraic quotient with any fractional part discarded; (see note 80)
Note 80 states (note that notes are non-normative):
80) This is often called truncation towards zero.
Point 4 goes on to state:
if the quotient a/b is representable in the type of the result, (a/b)*b + a%b is equal to a.
which can be shown to require the sign of a%b
to be the same as the sign of a
(when not zero).