NASM: Dividing large number with small one

NASM manual says on DIV:

• For DIV r/m32, EDX:EAX is divided by the given operand; the quotient is stored in EAX and the remainder in EDX.

What if EDX:EAX is a large number around 2⁵⁹ and the divider is 3? The quotient clearly cannot fit into EAX. Let's say I do not care about the remainder. I would like to have a best practice for doing the division.

Thinking about dividing the upper and lower 32 bits in separate steps. I think I could figure out some ugly result, but I would be interested in a good one. With a quick check for the case when EAX is likely to hold the quotient thus avoiding the complicated magic.

Solution: drhirsch's answer converted to NASM syntax:

; this divides edx:eax by ebx, even if the result is bigger than 2^32.
; result is in edx:eax, ecx,esi are used as spare registers
mov ecx, eax        开发者_运维问答   ;save lower 32 bit
mov eax, edx
xor edx, edx           ;now edx:eax contains 0:hi32 
div ebx
mov esi, eax           ;hi 32 bit of result, save to esi
mov eax, ecx           ;now edx:eax contains r:lo32, where r is the remainder
div ebx
mov edx, esi           ;restore hi32

---

This code is untested. It should calculate (d*2^32 + a)/b:

;this divides edx:eax by ebx, even if the result is bigger than 2^32.
;result is in edx:eax, ecx,esi are used as spare registers
;AT&T syntax.
  mov %eax, %ecx           ;save lower 32 bit
  mov %edx, %eax
  xor %edx, %edx           ;now edx:eax contains 0:hi32 
  div %ebx
  mov %eax, %esi           ;hi 32 bit of result
  mov %ecx, %eax           ;now edx:eax contains r:lo32, where r is the remainder
  div %ebx
  mov %esi, %edx           ;restore hi32