CODEWITHSUNDEEP
c++c++11language-lawyermodulo
RiaD
RiaD

Reputation: 47640

Does C++11 guarantee that the sign of the remainder is the sign of the dividend?

Is it guaranteed that in C++11, (-x) % m is negative and equal to (-x % m), where x and m are positive?

I know it's right on all machines I know.

Upvotes: 20

Views: 1603

Answers (2)

Christian Rau
Christian Rau

Reputation: 45948

In addition to Luchian's answer, this is the corresponding part from the C++11 standard:

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. For integral operands the / operator yields the algebraic quotient with any fractional part discarded; if the quotient a/b is representable in the type of the result, (a/b)*b + a%b is equal to a.

Which misses the last sentence. So the part

(a/b)*b + a%b is equal to a

Is the only reference to rely on, and that implies that a % b will always have the sign of a, given the truncating behaviour of /. So if your implementation adheres to the C++11 standard in this regard, the sign and value of a modulo operation is indeed perfectly defined for negative operands.

Upvotes: 11

Luchian Grigore
Luchian Grigore

Reputation: 258618

5.6 Multiplicative operators

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 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-defined (emphasis mine)

This is from C++03 though. :(

Upvotes: 7

Related Questions