user: 下面的汇编指令对应的C源代码是什么:

```asm
<predict_4x4_vr>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %rdx,%rax
  mov    %ecx,%edx
  lea    (%rcx,%rcx,1),%ecx
  mov    %esi,%r8d
  not    %r8d
  movzbl (%rdi,%r8,1),%r13d
  mov    $0xffffffff,%r8d
  movzbl (%rdi,%r8,1),%r12d
  lea    -0x1(%rsi),%r8d
  movzbl (%rdi,%r8,1),%r14d
  lea    -0x1(%rsi,%rsi,1),%r8d
  movzbl (%rdi,%r8,1),%ebx
  mov    %ebx,-0x4(%rsp)
  mov    %esi,%r8d
  neg    %r8d
  movzbl (%rdi,%r8,1),%ebx
  mov    $0x1,%r8d
  sub    %esi,%r8d
  movzbl (%rdi,%r8,1),%r10d
  mov    $0x2,%r8d
  sub    %esi,%r8d
  movzbl (%rdi,%r8,1),%r8d
  mov    $0x3,%r9d
  sub    %esi,%r9d
  movzbl (%rdi,%r9,1),%r15d
  lea    0x0(%r13,%rbx,1),%r11d
  movslq %ecx,%rsi
  lea    0x1(%r11),%edi
  sar    %edi
  mov    %dil,0x1(%rax,%rsi,1)
  mov    %dil,(%rax)
  lea    (%rbx,%r10,1),%r9d
  lea    0x1(%r9),%edi
  sar    %edi
  mov    %dil,0x2(%rax,%rsi,1)
  mov    %dil,0x1(%rax)
  lea    (%r10,%r8,1),%edi
  lea    0x1(%rdi),%ebp
  sar    %ebp
  mov    %bpl,0x3(%rax,%rsi,1)
  mov    %bpl,0x2(%rax)
  lea    0x1(%r8,%r15,1),%ebp
  sar    %ebp
  mov    %bpl,0x3(%rax)
  add    %edx,%ecx
  movslq %ecx,%rcx
  lea    (%r12,%r13,2),%ebp
  lea    0x2(%rbx,%rbp,1),%ebp
  sar    $0x2,%ebp
  mov    %bpl,0x1(%rax,%rcx,1)
  movslq %edx,%rdx
  mov    %bpl,(%rax,%rdx,1)
  add    %ebx,%r11d
  lea    0x2(%r10,%r11,1),%r11d
  sar    $0x2,%r11d
  mov    %r11b,0x2(%rax,%rcx,1)
  mov    %r11b,0x1(%rax,%rdx,1)
  add    %r10d,%r9d
  lea    0x2(%r8,%r9,1),%r9d
  sar    $0x2,%r9d
  mov    %r9b,0x3(%rax,%rcx,1)
  mov    %r9b,0x2(%rax,%rdx,1)
  add    %r8d,%edi
  lea    0x2(%r15,%rdi,1),%edi
  sar    $0x2,%edi
  mov    %dil,0x3(%rax,%rdx,1)
  lea    0x0(%r13,%r12,2),%edx
  lea    0x2(%r14,%rdx,1),%edx
  sar    $0x2,%edx
  mov    %dl,(%rax,%rsi,1)
  lea    (%r12,%r14,2),%edx
  mov    -0x4(%rsp),%ebx
  lea    0x2(%rbx,%rdx,1),%edx
  sar    $0x2,%edx
  mov    %dl,(%rax,%rcx,1)
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
```

assistant: ```c
void predict_4x4_vr(unsigned char *src, unsigned int src_stride, unsigned char *dst, unsigned int dst_stride, int neighbour) {
  const int dst_row_0 = 0;
  const int dst_row_1 = dst_stride;
  const int dst_row_2 = dst_stride * 2;
  const int dst_row_3 = dst_row_1 + dst_row_2;
  const int lt = src[-1 - src_stride];
  const int l0 = src[-1 + 0 * src_stride];
  const int l1 = src[-1 + 1 * src_stride];
  const int l2 = src[-1 + 2 * src_stride];
  const int l3 = src[-1 + 3 * src_stride];
  const int t0 = src[0 - 1 * src_stride];
  const int t1 = src[1 - 1 * src_stride];
  const int t2 = src[2 - 1 * src_stride];
  const int t3 = src[3 - 1 * src_stride];
  dst[dst_row_0 + 0] = dst[dst_row_2 + 1] = (lt + t0 + 1) >> 1;
  dst[dst_row_0 + 1] = dst[dst_row_2 + 2] = (t0 + t1 + 1) >> 1;
  dst[dst_row_0 + 2] = dst[dst_row_2 + 3] = (t1 + t2 + 1) >> 1;
  dst[dst_row_0 + 3] = (t2 + t3 + 1) >> 1;
  dst[dst_row_1 + 0] = dst[dst_row_3 + 1] = (l0 + 2 * lt + t0 + 2) >> 2;
  dst[dst_row_1 + 1] = dst[dst_row_3 + 2] = (lt + 2 * t0 + t1 + 2) >> 2;
  dst[dst_row_1 + 2] = dst[dst_row_3 + 3] = (t0 + 2 * t1 + t2 + 2) >> 2;
  dst[dst_row_1 + 3] = (t1 + 2 * t2 + t3 + 2) >> 2;
  dst[dst_row_2 + 0] = (lt + 2 * l0 + l1 + 2) >> 2;
  dst[dst_row_3 + 0] = (l0 + 2 * l1 + l2 + 2) >> 2;
}
```