CSE3101 Fall 2005 Exam #2, open book, open notes.  Name __________________

1.  Write an equivalent function in assembler (30 pts):

  double f(int a, double x) {
    return (a/x) * 3.0 - 2.0;
  }


  ; ANSWER
  two   dq 2.0  ; or two real8 2.0, could also be in data segment
  three dq 3.0
_f:
  fild dword ptr [esp+4]  ; a (need to specify operand size)
  fdiv qword ptr [esp+8]  ; x
  fmul [three]            ; ("qword ptr" or "real8 ptr" is optional here)
  fsub [two]
  ret                     ; result in st(0)


2.  The code below adds two vectors of length n.  Write an equivalent
function in assembler.  You may assume n > 0  (30 pts).

  void add(double *a, double *b, double *sum, int n) {
    for (int i=0; i<n; ++i)
      sum[i] = a[i] + b[i];
  }


  ; ANSWER
_add:
  push ebx          ; don't need to save eax, ecx, edx
  mov eax, [esp+8]  ; a (register choice is somewhat arbitrary)
  mov ebx, [esp+12] ; b
  mov edx, [esp+16] ; sum
  mov ecx, [esp+20] ; n, counts down to 1
L1:
  fld  qword ptr [eax+ecx*8-8]  ; a[n-1]
  fadd qword ptr [ebx+ecx*8-8]  ; + b[n-1]
  fstp qword ptr [edx+ecx*8-8]
  dec ecx  ; or sub ecx,1 or loop L1
  jnz L1
  pop ebx
  ret


3.  When is MOVNTQ faster than MOVQ? (10 pts)

  ANSWER: when writing to memory that is not in cache and would otherwise
  be flushed from cache before being used again, such as when filling
  or copying a large array.


4.  The following subtracts n 16-bit unsigned elements of vector b
from vector a.  If an element would be negative it is set to 0 instead.
Write an equivalent function using MMX to compute 4 elements in parallel.
You may assume n is a positive multiple of 4  (30 pts).

  void sub(unsigned short *a, unsigned short *b, int n) {
    for (int i=0; i<n; ++i) {
      if (a[i] < b[i])
        a[i] = 0;
      else
        a[i] -= b[i];
    }
  }


  ; ANSWER
_sub:
  mov eax, [esp+4]  ; a
  mov edx, [esp+8]  ; b
  mov ecx, [esp+12] ; n, counts by 4 down to 4
L1:
  movq mm0, [eax+ecx*2-8]     ; a[n-4]...a[n-1]
  psubusw mm0, [edx+ecx*2-8]  ; - b[n-4]...b[n-1], saturates to minimum of 0
  movq mm0, [eax+ecx*2-8]     ; save, don't use movntq because memory is already cached
  sub ecx, 4
  ja L1  ; or jg or jne or jnz
  emms   ; don't forget
  ret