Modulo operator with negative values [duplicate]

This question already has answer开发者_如何转开发s here: Why does C++ output negative numbers when using modulo? (4 answers) Closed 3 years ago.

Why do such operations:

std::cout << (-7 % 3) << std::endl;
std::cout << (7 % -3) << std::endl;

give different results?

-1
1

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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)

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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

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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.