Remainder Operator(%) in C++
What is value of
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
b is negative is undefined.
But in C++ 11, the result is well-defined, the standard says (emphasis added by
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
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
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
b is -1 and 1 respectively.
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