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.