# Remainder Operator(%) in C++

··294 words·2 mins·

What is value of `a` and `b` in the following statement according to the rules of C++ 11?

``````int a = -5 % 2;
int b = 5 % -2;
``````

If you can not clearly tell the result, then you should update your knowledge about % operator in C++.

# The rules for remainder operator#

Some people call `%` modulo operator, but its real name is remainder operator. Rules about % operator differs between C++ 11 and pre-C++ 11 era. In pre-C++ 11 era, the result of `a%b` when `a` or `b` is negative is undefined. But in C++ 11, the result is well-defined, the standard says (emphasis added by me):

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.

It basically says that `a == (a/b)*b + (a%b)` and the result of `a/b` is truncated toward zero. With these two rules, the result of `a%b` can be calculated without any ambiguity:

``````a%b = a - (a/b)*b
``````

# A real example#

Take the calculation of `a` and `b` in the first paragraph as an example,

``````-5/2 ==> -2
-5%2 = -5 - (-2*2) = -1

5/-2 = -2
5%-2 = 5 - (-2*-2) = 1
``````

So, in C++ 11, the value for `a` and `b` is -1 and 1 respectively.

# References#

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