**Parallel Programe
HW1:**

**Source:**

void test(float restrict a, float restrict b, float restrict c, int N) {
    builtinassume(N == 1024);
    
    a = (float )builtinassumealigned(a, 16);
    b = (float )builtinassumealigned(b, 16);
    c = (float )builtinassumealigned(c, 16);
  
    for (int i=0; i<I; i++) {
        for (int j=0; j<N; j++) {
            c[j] = a[j] + b[j];
        }
    }
}

Build Command:

make clean && make VECTORIZE=1 AVX2=1 && ./test_auto_vectorize -t 1

**Q1:**

Fix the code to make sure it uses aligned moves for the best performance
Hint: we want to see vmovaps rather than vmovups.

**Answer:**

void test(float restrict a, float restrict b, float restrict c, int N) {
  builtinassume(N == 1024);
  a = (float )builtinassumealigned(a, 32);
  b = (float )builtinassumealigned(b, 32);
  c = (float )builtinassumealigned(c, 32);
  
  for (int i=0; i<I; i++) {
    for (int j=0; j<N; j++) {
      c[j] = a[j] + b[j];
    }
  }
}

Check Command:
diff assembly/test1.vec.restr.align.s assembly/test1.vec.restr.align.avx2.s


**Q2:**

What speedup does the vectorized code achieve over the unvectorized code? 
What additional speedup does using -mavx2 give (AVX2=1 in the Makefile)? 
You may wish to run this experiment several times and take median elapsed times; 
you can report answers to the nearest 100% (e.g., 2×, 3×, etc). 
What can you infer about the bit width of the default vector registers on the PP machines? 
What about the bit width of the AVX2 vector registers.

Hint: Aside from speedup and the vectorization report, the most relevant information is that the data type for each array is float.

**Answer:**

What speedup does the vectorized code achieve over the unvectorized code?
unvectorized code is slow than the vectorized code.
> Ans: about 3.5X

What additional speedup does using -mavx2 give (AVX2=1 in the Makefile)?
using -mavx2 is faster than the vectorized code.
> Ans: about 2x

What can you infer about the bit width of the default vector registers on the PP machines?
> Ans:
movaps	(%rbx,%rdx,4), %xmm0
movaps	16(%rbx,%rdx,4), %xmm1
addps	(%r15,%rdx,4), %xmm0
addps	16(%r15,%rdx,4), %xmm1
movaps	%xmm0, (%r14,%rdx,4)
movaps	%xmm1, 16(%r14,%rdx,4)
movaps	32(%rbx,%rdx,4), %xmm0
movaps	48(%rbx,%rdx,4), %xmm1
addps	32(%r15,%rdx,4), %xmm0
addps	48(%r15,%rdx,4), %xmm1
差異都是16

What about the bit width of the AVX2 vector registers?
> Ans:
vmovups	(%rbx,%rdx,4), %ymm0
vmovups	32(%rbx,%rdx,4), %ymm1
vmovups	64(%rbx,%rdx,4), %ymm2
vmovups	96(%rbx,%rdx,4), %ymm3
vaddps	(%r15,%rdx,4), %ymm0, %ymm0
vaddps	32(%r15,%rdx,4), %ymm1, %ymm1
vaddps	64(%r15,%rdx,4), %ymm2, %ymm2
vaddps	96(%r15,%rdx,4), %ymm3, %ymm3
vmovups	%ymm0, (%r14,%rdx,4)
vmovups	%ymm1, 32(%r14,%rdx,4)
vmovups	%ymm2, 64(%r14,%rdx,4)
vmovups	%ymm3, 96(%r14,%rdx,4)
差異都是32

**Q3:**

Provide a theory for why the compiler is generating dramatically different assembly.

**Answer:**

Theory:Basic Block
程式的流程被改變了
原先是先賦值,再來判斷去決定是否修改C得值,等於是至少會做一次賦值的動作
修改後則是利用判斷大小來決定賦值的動作