Modulo operator with negative values
C++C++ Problem Overview
Why do such operations:
std::cout << (-7 % 3) << std::endl;
std::cout << (7 % -3) << std::endl;
give different results?
-1
1
C++ Solutions
Solution 1 - C++
From ISO14882:2011(e) 5.6-4:
> 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. 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.
The rest is basic math:
(-7/3) => -2
-2 * 3 => -6
so a%b => -1
(7/-3) => -2
-2 * -3 => 6
so a%b => 1
Note that
> If both operands are nonnegative then the remainder is nonnegative; if > not, the sign of the remainder is implementation-defined.
from ISO14882:2003(e) is no longer present in ISO14882:2011(e)
Solution 2 - C++
a % b
in c++ default:
(-7/3) => -2
-2 * 3 => -6
so a%b => -1
(7/-3) => -2
-2 * -3 => 6
so a%b => 1
in python:
-7 % 3 => 2
7 % -3 => -2
in c++ to python:
(b + (a%b)) % b
Solution 3 - C++
The sign in such cases (i.e when one or both operands are negative) is implementation-defined. The spec says in §5.6/4 (C++03),
>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.
That is all the language has to say, as far as C++03 is concerned.