**Integer division** in modern C always rounds towards zero. For example:

```
9/2 => 4
-9/2 => -4 (not -5)
```

This was was not always so. In early C versions and in the C89 standard, positive integer division rounded to zero, but result of negative integer division was implementation dependent!

From **§3.3.5 Multiplicative operators** of the **C89 standard**:

When integers are divided and the division is inexact, if both operands are positive the result of the

`/`

operator is the largest integer less than the algebraic quotient and the result of the`%`

operator is positive. If either operand is negative, whether the result of the`/`

operator is the largest integer less than the algebraic quotient or the smallest integer greater than the algebraic quotient is implementation-defined, as is the sign of the result of the`%`

operator. If the quotient`a/b`

is representable, the expression`(a/b)*b + a%b`

shall equal`a`

.

The same appears in the first edition of **The C Programming Language** book by Kernighan and Ritchie.

This implementation-defined behavior was fixed by the C99 standard which defined that integer division always rounds towards zero.

From **§6.5.5 Multiplicative operators** of the **C99 standard**:

When integers are divided, the result of the

`/`

operator is the algebraic quotient with any fractional part discarded.

Thanks to Arch for pointing this out.