On a IA-32 architecture how can i divide a signed number by 3 (e.g.) a value stored in 2 registers, edx:eax (a 64-bit value). I want to divide the whole value (64-bits) by 3, not only 32-bits, and store it in 2 registers.
I'm assuming this can only be done using shifts operations since imul only works for multiplying 32-bits numbers. But only found solutions for dividing by 2^n numbers.
How can i achieve this?
You can divide any length number by a 32 bit number with successive divides, using the remainder of the prior divide as the most significant 32 bits of the dividend for the next divide, similar to long hand division
Note I need to fix this code to handle negative divisors, but it should work with positive divisors and signed dividend.
Note this code rounds towards negative infinity: -10/3 : quotient = -4, remainder = +2. To handle negative divisors, the code could negate both divisor and dividend, then negate the remainder after.
mov ecx,000000003h ;ecx = signed dvsr (must be positive)
mov edi,0fedcba98h ;edi:esi = signed dvnd
mov esi,076543210h
;; inputs
mov eax,edi ;eax = upper 32 bits dvnd
cdq ; sign-extend that into EDX:EAX
idiv ecx
test edx,edx ;br if sign rmdr == sign dvsr
jns short div0
dec eax ;dec quot
add edx,ecx ;rem += dvsr
div0: mov edi,eax ;edi = upper 32 bits quot
mov eax,esi ;eax = lower 32 bits dvnd
div ecx
mov esi,eax ;esi = lower 32 bits quot
; ;edx = remainder