factorial-recursion-asm-walkthrough

*C function*
int fact(int n){
  if (n <= 1) return 1;
  return n * fact(n-1);
}
--
*NASM code*
default rel
global _main
section .data
number: dd 5

section .text
_main:  mov eax, 1
        mov edi, [number]
        call .fact
.fact:  cmp edi, 2
        jl .basecase
        push rbx
        mov ebx, edi
        lea edi, [rbx - 1]
        call .fact
        imul eax, ebx
        pop rbx

.basecase:
        ret
--

*Step-through*
Sample function call: 
fact(5);

edi = 5
1st pass:
eax = 1
cmp edi, 2
islesser? No -> move on
push rbx
mov ebx, edi => ebx = 5
lea edi, rbx - 1 => edi = 4
call fact

ebx=5, edi=4
2nd pass:
eax=1
cmp edi, 2 // edi < 2 No, move on
push rbx
mov ebx, edi => ebx = 4
lea edi, rbx-1 => edi = 3
call fact

ebx=4, edi=3
3rd pass:
eax=1
cmp edi, 2 // edi < 2 No, move on
push rbx
mov ebx, edi => ebx = 3
lea edi, rbx-1 => edi = 2
call fact

ebx=3, edi=2
4th pass:
eax=1
cmp edi, 2 // edi<2 No, move on
push rbx
mov ebx, edi => ebx=2
lea edi, rbx-1 => edi = 1
call fact

ebx=2, edi=1
5th pass:
eax=1
cmp edi,2 // edi < 2, Yes
ret to 4th pass with eax=1
imul eax, ebx 1 * 2 = 2
pop rbx
ret to 3rd pass with eax=2
imul eax, ebx => 2 * 3 = 6
pop rbx
ret to 2nd pass with eax=6
imul eax, ebx => 6 * 4 = 24
pop rbx
ret to 1st pass with eax=24
imul eax, ebx => 24 * 5 = 120

ret with eax=120