Integer division rounding with negatives in C++

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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).

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