C++ – How Code Alignment Dramatically Affects Performance

Today I have found sample code which slowed down by 50%, after adding some unrelated code. After debugging I have figured out the problem was in the loop alignment.
Depending of the loop code placement there is different execution time e.g.:

I didn't expect previously that code alignment may have such a big impact. And I thought my compiler is smart enough to align the code correctly.

What exactly cause such a big difference in execution time ? (I suppose some processor architecture details).

The test program I have compiled in Release mode with Visual Studio 2019 and run it on Windows 10.
I have checked the program on 2 processors: i7-8700k (the results above), and on intel i5-3570k but the problem does not exist there and the execution time is always about 1250us.
I have also tried to compile the program with clang, but with clang the result is always ~1500us (on i7-8700k).

My test program:

#include <chrono>
#include <iostream>
#include <intrin.h>
using namespace std;

template<int N>
__forceinline void noops()
{
    __nop(); __nop(); __nop(); __nop(); __nop(); __nop(); __nop(); __nop(); __nop(); __nop(); __nop(); __nop(); __nop(); __nop(); __nop(); __nop();
    noops<N - 1>();
}
template<>
__forceinline void noops<0>(){}

template<int OFFSET>
__declspec(noinline) void SumHorizontalLine(const unsigned char* __restrict src, int width, int a, unsigned short* __restrict dst)
{
    unsigned short sum = 0;
    const unsigned char* srcP1 = src - a - 1;
    const unsigned char* srcP2 = src + a;

    //some dummy loop,just a few iterations
    for (int i = 0; i < a; ++i)
        dst[i] = src[i] / (double)dst[i];

    noops<OFFSET>();
    //the important loop
    for (int x = a + 1; x < width - a; x++)
    {
        unsigned char v1 = srcP1[x];
        unsigned char v2 = srcP2[x];
        sum -= v1;
        sum += v2;
        dst[x] = sum;
    }

}

template<int OFFSET>
void RunTest(unsigned char* __restrict src, int width, int a, unsigned short* __restrict dst)
{
    double minTime = 99999999;
    for(int i = 0; i < 20; ++i)
    {
        auto start = chrono::steady_clock::now();

        for (int i = 0; i < 1024; ++i)
        {
            SumHorizontalLine<OFFSET>(src, width, a, dst);
        }

        auto end = chrono::steady_clock::now();
        auto us = chrono::duration_cast<chrono::microseconds>(end - start).count();
        if (us < minTime)
        {
            minTime = us;
        }
    }

    cout << OFFSET << " : " << minTime << " us" << endl;
}

int main()
{
    const int width = 2048;
    const int x = 3;
    unsigned char* src = new unsigned char[width * 5];
    unsigned short* dst = new unsigned short[width];
    memset(src, 0, sizeof(unsigned char) * width);
    memset(dst, 0, sizeof(unsigned short) * width);

    while(true)
    RunTest<1>(src, width, x, dst);
}

To verify different alignment, just recompile the program and change RunTest<0> to RunTest<1> etc.
Compiler always align the code to 16bytes. In my test code I just insert additional nops to move the code a bit more.

Assembly code generated for the loop with OFFSET=1 (for other offset only the amount of npads is different):

  0007c 90       npad    1
  0007d 90       npad    1
  0007e 49 83 c1 08  add     r9, 8
  00082 90       npad    1
  00083 90       npad    1
  00084 90       npad    1
  00085 90       npad    1
  00086 90       npad    1
  00087 90       npad    1
  00088 90       npad    1
  00089 90       npad    1
  0008a 90       npad    1
  0008b 90       npad    1
  0008c 90       npad    1
  0008d 90       npad    1
  0008e 90       npad    1
  0008f 90       npad    1
$LL15@SumHorizon:

; 25   : 
; 26   :    noops<OFFSET>();
; 27   : 
; 28   :    for (int x = a + 1; x < width - a; x++)
; 29   :    {
; 30   :        unsigned char v1 = srcP1[x];
; 31   :        unsigned char v2 = srcP2[x];
; 32   :        sum -= v1;

  00090 0f b6 42 f9  movzx   eax, BYTE PTR [rdx-7]
  00094 4d 8d 49 02  lea     r9, QWORD PTR [r9+2]

; 33   :        sum += v2;

  00098 0f b6 0a     movzx   ecx, BYTE PTR [rdx]
  0009b 48 8d 52 01  lea     rdx, QWORD PTR [rdx+1]
  0009f 66 2b c8     sub     cx, ax
  000a2 66 44 03 c1  add     r8w, cx

; 34   :        dst[x] = sum;

  000a6 66 45 89 41 fe   mov     WORD PTR [r9-2], r8w
  000ab 49 83 ea 01  sub     r10, 1
  000af 75 df        jne     SHORT $LL15@SumHorizon

; 35   :    }
; 36   : 
; 37   : }

  000b1 c3       ret     0
??$SumHorizontalLine@$00@@YAXPEIBEHHPEIAG@Z ENDP    ; SumHorizont