Fast Division on GCC/ARM

Question

As far as I know most compilers will do fast division by multiplying and then bit shifting to the right. For instance, if you check this SO thread it says that when you ask the Microsoft compiler to do division by 10 it will multiply the dividend by 0x1999999A (which is 2^32/10) and then divide the result by 2^32 (using 32 shifts to the right).

(Editor's note: that linked answer was wrong until. Compilers don't do that because it's not exact for all inputs. Compilers do multiply and shift, but with a more involved way of determining the magic constant and shift counts: Why does GCC use multiplication by a strange number in implementing integer division?)

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So far so good.

Once I tested the same division by 10 on ARM using GCC, though, the compiler did something slightly different. First it multiplied the dividend by 0x66666667 (2^34/10), then it divided the result by 2^34. Thus far it's the same as Microsoft, except using a higher multiplier. After that, however, it subtracted (dividend/2^31) from the result.

My question: why on the ARM version there's that extra subtraction? Can you give me an numeric example where without that subtraction the result will be wrong?

If you want to check the generated code, it's below (with my comments):

        ldr     r2, [r7, #4] @--this loads the dividend from memory into r2
        movw    r3, #:lower16:1717986919 @--moves the lower 16 bits of the constant 
        movt    r3, #:upper16:1717986919 @--moves the upper 16 bits of the constant
        smull   r1, r3, r3, r2 @--multiply long, put lower 32 bits in r1, higher 32 in r3
        asr     r1, r3, #2 @--r3>>2, then store in r1 (effectively >>34, since r3 was higher 32 bits of multiplication)
        asr     r3, r2, #31 @--dividend>>31, then store in r3
        rsb     r3, r3, r1 @--r1 - r3, store in r3
        str     r3, [r7, #0] @--this stores the result in memory (from r3) 

Answer 1

After that, however, it subtracted (dividend/2^31) from the result.

Actually, it subtracts dividend >> 31, which is -1 for negative dividend, and 0 for non-negative dividend, when right-shifting negative integers is arithmetic right-shift (and int is 32 bits wide).

0x6666667 = (2^34 + 6)/10

So for x < 0, we have, writing x = 10*k + r with -10 < r <= 0,

0x66666667 * (10*k+r) = (2^34+6)*k + (2^34 + 6)*r/10 = 2^34*k + 6*k + (2^34+6)*r/10

Now, arithmetic right shift of negative integers yields the floor of v / 2^n, so

(0x66666667 * x) >> 34

results in

k + floor((6*k + (2^34+6)*r/10) / 2^34)

So we need to see that

-2^34 < 6*k + (2^34+6)*r/10 < 0

The right inequality is easy, both k and r are non-positive, and not both are 0.

For the left inequality, a bit more analysis is needed.

r >= -9

so the absolute value of (2^34+6)*r/10 is at most 2^34+6 - (2^34+6)/10.

|k| <= 2^31/10,

so |6*k| <= 3*2^31/5.

And it remains to verify that

6 + 3*2^31/5 < (2^34+6)/10
1288490194   < 1717986919

Yup, true.