How can I convert this assembly code (x64) to C?

Question

I have the following assembly (x64) function:

400700: sub  rsp,0x18
400704: mov  QWORD PTR [rsp+0x8],rdi
400709: cmp  QWORD PTR [rsp+0x8],0x0
40070f: jne  400718 <bar+0x18>
400711: mov  eax,0x0
400716: jmp  400734 <bar+0x34>
400718: mov  rax,QWORD PTR [rsp+0x8]
40071d: sub  rax,0x1
400721: mov  rdi,rax
400724: call 400700 <bar>
400729: mov  rdx,rax
40072c: mov  rax,QWORD PTR [rsp+0x8]
400731: add  rax,rdx
400734: add  rsp,0x18
400738: ret

I've figured out that the beginning part is just the header for any assembly program. Also, I think that the two QWORD are either from longs or characters. I'm guessing that the jne jump is probably a conditional statement, but I'm not too sure. It looks like the function is recursive since it has a line that calls "bar" again.

My guess for the overall function is that it takes two longs and recursively adds them (Fibonacci, maybe?). I don't have much reasoning for this guess though, and I could be incorrect.

Could anyone please help me solve this problem? Or, if there's a way for me to pass in parameters (I don't even know what kind of argument the function takes) and see the output, then I can try to backtrack and figure out what the program is doing.

Thanks

Answer 1

This function computes (in a recursive manner) the sum of the n first integers:

     400700: sub  rsp,0x18                 ; Align the stack
     400704: mov  QWORD PTR [rsp+0x8],rdi  ; Store first argument on stack
     400709: cmp  QWORD PTR [rsp+0x8],0x0  ; test (n == 0)
  +--40070f: jne  400718 <bar+0x18>        ; jump to 400718 if (n != 0)
  |  400711: mov  eax,0x0                  ; set return code to zero
+-|--400716: jmp  400734 <bar+0x34>        ; jump to function end at 400734
| +->400718: mov  rax,QWORD PTR [rsp+0x8]  ; Get 'n' in rax
|    40071d: sub  rax,0x1                  ; n = n - 1
|    400721: mov  rdi,rax                  ; Load 'n - 1' as first argument
|    400724: call 400700 <bar>             ; recursive call to 'bar(n - 1)'
|    400729: mov  rdx,rax                  ; Load bar(n-1) return code in rdx
|    40072c: mov  rax,QWORD PTR [rsp+0x8]  ; Get current 'n' in rax
|    400731: add  rax,rdx                  ; rax = n + bar(n - 1)
+--->400734: add  rsp,0x18                 ; Restore the stack
     400738: ret                           ; return n + bar(n - 1)

Final code:

bar (long long int n)
{
  return (n) ? n + bar(n - 1) : 0;
}

Frankly, I would have used the well known formula n*(n+1)/2...