Convert function to Arm Neon - c

Question

I'm a beginner in Arm Neon, and I'm trying to vectorise this loop
float ans=0.0;
for (i=0; i<numdims; i++)
ans += (pt1[i]-pt2[i]) * (pt1[i]-pt2[i]);
I'm trying to convert this function in Neon with prefetch instruction and loop-unrolling
int iter= numdims/4*4;
float result[3];
float ans=0.0;
asm volatile(
"mov x1, #0\n\t"
"mov x2, %[pt1]\n\t"
"mov x3, %[pt2]\n\t"
"movi v3.4s, #0\n\t"
".loop_neon%=:\n\t"
"prfm PLDL1STRM, [x2, #64]\n\t"
"prfm PLDL1STRM, [x3, #64]\n\t"
"ldr q1, [x2, #16]\n\t"
"ldr q2, [x3, #16]\n\t"
"fsub v4.4s, v1.4s, v2.4s\n\t"
"fmla v3.4s, v4.4s, v4.4s\n\t"
"add x1,x1, #16\n\t"
"cmp x1, %[iter]\n\t"
"b.lt .loop_neon%=\n\t"
"str q3, [%[result]]\n\t"
:
: [iter] "r" (iter),[pt1] "r" (pt1),[pt2] "r" (pt2), [result] "r" (result)
: "x1","x2","x3","memory","v0","v1","v2","v3","v4"
);
ans = result[0] + result[1] + result[2] + result[3];
//final iterations of the loop
for (int i=iter; i<numdims; i++)
ans += (pt1[i]-pt2[i]) * (pt1[i]-pt2[i]);
This code works but the output is not correct

Answer 1

Short answer: add x1, x1, #4
Your code is far from optimal:
there are lots of pipeline hazards. unroll deeper
you should always count down the loop counter
you should avoid unnecessary memory access (result)
you should avoid unnecessary mov operations
Provided iter is a multiple of 16, the code below is suggested:
.func
// extern float sumDiffSquare(float *pA, float *pB, uint32_t length);
// assert(length >= 16);
// assert(length & 15 == 0);
pA .req x0
pB .req x1
length .req x2
sumDiffSqare:
movi v0.16b, #0
.balign 64
1:
ldp q16, q17, [pA], #32
ldp q20, q21, [pB], #32
ldp q18, q19, [pA], #32
ldp q22, q23, [pB], #32
subs length, length, #16
fsub v16.4s, v20.4s, v16.4s
fsub v17.4s, v21.4s, v17.4s
fsub v18.4s, v22.4s, v18.4s
fsub v19.4s, v23.4s, v19.4s
fmla v0.4s, v16.4s, v16.4s
fmla v0.4s, v17.4s, v17.4s
fmla v0.4s, v18.4s, v18.4s
fmla v0.4s, v19.4s, v19.4s
b.gt 1b
faddp v0.4s, v0.4s, v0.4s
faddp v0.2s, v0.2s, v0.2s
ret
.endfunc