Divide by 10 in assembly not giving correct result

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I was trying to implement division by 10 in arm assembly. I followed the method mentioned in the first answer to the question at ARM DIVISION HOW TO DO IT?

This is my code

MOV     r2, #10         
LDR     r3,=0x1999999A   ; 1999999A == (2^32 / 10) + 1
UMULL   r9,r3,r5,r3    ;divide an integer value in r5 by 10   
MOV     r6, r9,LSR r2  ; r6 has the quotient

But this code is not giving the correct result.For example, If I give 0x0000000B in r5, after these steps value in r6 is 0x00066666.

Am I doing anything wrong?

assembly
arm
integer-division
asked on Stack Overflow Apr 14, 2014 by User1234321232 • edited May 23, 2017 by Community

2 Answers

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uint32(x / 10) = uint64(x * 0xcccccccd) >> 35

E.g. the following code will divide unsigned r5 by 10 with the quotient in r0:

ldr    r2,=0xcccccccd
umull  r0,r1,r5,r2
mov    r0,r1,lsr #3
answered on Stack Overflow Apr 14, 2014 by apangin
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There is a rather good writeup of division by constant http://thinkingeek.com/2013/08/11/arm-assembler-raspberry-pi-chapter-15/ It has a python script to generate the magic numbers and the code for any constant. For 10 it's

/* r0 contains the argument to be divided by 10 */
ldr r1, .Lu_magic_number_10 /* r1 ← magic_number */
umull r1, r2, r1, r0   /* r1 ← Lower32Bits(r1*r0). r2 ← Upper32Bits(r1*r0) */
mov r0, r2, LSR #3     /* r0 ← r2 >> 3 */
bx lr                  /* leave function */
.align 4
  .Lu_magic_number_10: .word 0xcccccccd

And the arm website has the code on it, but i found it here now http://www.sciencezero.org/index.php?title=ARM:_Division_by_10

Note that this even manages without the later introduced 64bit multiplication.

answered on Stack Overflow Nov 12, 2015 by Torsten

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