What is the behavior of integer division?

Solution 1:

Will result always be the floor of the division? What is the defined behavior?

Not quite. It rounds toward 0, rather than flooring.

6.5.5 Multiplicative operators

6 When integers are divided, the result of the / operator is the algebraic quotient with any fractional part discarded.⁸⁸⁾ If the quotient a/b is representable, the expression (a/b)*b + a%b shall equal a.

and the corresponding footnote:

88. This is often called ‘‘truncation toward zero’’.

Of course two points to note are:

3 The usual arithmetic conversions are performed on the operands.

and:

5 The result of the / operator is the quotient from the division of the first operand by the second; the result of the % operator is the remainder. In both operations, if the value of the second operand is zero, the behavior is undefined.

[Note: Emphasis mine]

Solution 2:

Dirkgently gives an excellent description of integer division in C99, but you should also know that in C89 integer division with a negative operand has an implementation-defined direction.

From the ANSI C draft (3.3.5):

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.

So watch out with negative numbers when you are stuck with a C89 compiler.

It's a fun fact that C99 chose truncation towards zero because that was how FORTRAN did it. See this message on comp.std.c.

Solution 3:

Yes, the result is always truncated towards zero. It will round towards the smallest absolute value.

-5 / 2 = -2
 5 / 2 =  2

For unsigned and non-negative signed values, this is the same as floor (rounding towards -Infinity).