Why is such complex code emitted for dividing a signed integer by a power of two?

Question

When I compile this code with VC++10:

DWORD ran = rand();
return ran / 4096;

I get this disassembly:

299: {
300:    DWORD r         


        

Answer 1

the "extra manipulations" compensate for the fact that arithmetic right-shift rounds the result toward negative infinity, whereas division rounds the result towards zero.

For example, -1 >> 1 is -1, whereas -1/2 is 0.

Answer 2

From the C standard:

When integers are divided, the result of the / operator is the algebraic quotient with any fractional part discarded.105) If the quotient a/b is representable, the expression (a/b)*b + a%b shall equal a; otherwise, the behavior of both a/b and a%b is undefined.

It's not hard to think of examples where negative values for a don't follow this rule with pure arithmetic shift. E.g.

(-8191) / 4096 -> -1
(-8191) % 4096 -> -4095

which satisfies the equation, whereas

(-8191) >> 12 -> -2 (assuming arithmetic shifting)

is not division with truncation, and therefore -2 * 4096 - 4095 is most certainly not equal to -8191.

Note that shifting of negative numbers is actually implementation-defined, so the C expression (-8191) >> 12 does not have a generally correct result as per the standard.

Answer 3

The reason is that unsigned division by 2^n can be implemented very simply, whereas signed division is somewhat more complex.

unsigned int u;
int v;

u / 4096 is equivalent to u >> 12 for all possible values of u.

v / 4096 is NOT equivalent to v >> 12 - it breaks down when v < 0, as the rounding direction is different for shifting versus division when negative numbers are involved.