user: 这段汇编代码用C语言表示是什么样的？

```asm
<gf8_muladd_CB>:
  endbr64
  test   %edx,%edx
  je     L19
  push   %r15
  mov    %rdi,%rax
  lea    (%rdx,%rdx,1),%edi
  mov    %rsi,%rcx
  push   %r14
  mov    %rdi,%rsi
  shl    $0x2,%rdi
  push   %r13
  lea    0x0(,%rdx,8),%r13d
  push   %r12
  mov    %edx,%r12d
  sub    %edx,%r13d
  push   %rbp
  shl    $0x2,%r12
  shl    $0x2,%r13
  push   %rbx
  mov    %r12d,%r14d
  shl    $0x2,%r14
  mov    %rdi,-0x10(%rsp)
  lea    (%rsi,%rdx,1),%edi
  mov    %rdi,%rsi
  shl    $0x2,%rdi
  add    %esi,%esi
  mov    %rdi,-0x28(%rsp)
  lea    (%rdx,%rdx,4),%edi
  lea    0x0(,%rsi,4),%r15
  lea    (%rax,%r12,1),%rsi
  mov    %rsi,-0x8(%rsp)
  lea    0x0(,%rdi,4),%r10
  nopl   (%rax)
L4:
  mov    (%rax,%r14,1),%r9d
  mov    (%rax,%r10,1),%r8d
  add    $0x4,%rcx
  add    $0x4,%rax
  mov    -0x28(%rsp),%rbx
  mov    -0x4(%rax,%r13,1),%ebp
  mov    -0x4(%rax,%r13,1),%r11d
  mov    -0x4(%rax),%edi
  mov    -0x4(%rax,%rbx,1),%ebx
  xor    %r9d,%ebp
  xor    -0x4(%rcx),%r9d
  xor    %r8d,%r11d
  mov    -0x4(%rax,%r15,1),%esi
  mov    %ebx,%edx
  mov    %r11d,-0x14(%rsp)
  mov    -0x4(%rax,%r12,1),%r11d
  xor    %ebp,%edx
  mov    %ebx,-0x18(%rsp)
  mov    %edx,%ebx
  mov    -0x10(%rsp),%rdx
  xor    %edi,%r11d
  mov    %r11d,-0x1c(%rsp)
  xor    %ebp,%r11d
  xor    -0x4(%rax,%rdx,1),%r11d
  xor    %r11d,%r9d
  mov    %r9d,-0x4(%rax)
  mov    -0x4(%rcx,%r12,1),%r9d
  xor    %ebx,%r9d
  xor    %r11d,%r9d
  mov    %r9d,-0x4(%rax,%r12,1)
  xor    -0x4(%rcx,%rdx,1),%ebx
  mov    %ebx,-0x4(%rax,%rdx,1)
  mov    -0x28(%rsp),%rbx
  mov    -0x18(%rsp),%edx
  xor    -0x4(%rcx,%rbx,1),%r8d
  xor    %r11d,%r8d
  xor    %esi,%edx
  mov    %r8d,-0x4(%rax,%rbx,1)
  mov    -0x14(%rsp),%ebx
  xor    -0x4(%rcx,%r14,1),%edx
  xor    %ebx,%edx
  mov    %edx,-0x4(%rax,%r14,1)
  mov    -0x4(%rcx,%r10,1),%edx
  xor    %esi,%edx
  xor    %ebp,%edx
  mov    %edx,-0x4(%rax,%r10,1)
  xor    -0x4(%rcx,%r15,1),%edi
  xor    %ebx,%edi
  mov    %edi,-0x4(%rax,%r15,1)
  xor    -0x4(%rcx,%r13,1),%esi
  xor    -0x1c(%rsp),%esi
  mov    %esi,-0x4(%rax,%r13,1)
  cmp    -0x8(%rsp),%rax
  jne    L4
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax,%rax,1)
L19:
  ret
```

assistant: ```c
void gf8_muladd_CB(uint8_t *out, uint8_t *in, unsigned int width) {
  unsigned int i;
  uint64_t *in_ptr = (uint64_t *)in;
  uint64_t *out_ptr = (uint64_t *)out;
  for (i = 0; i < width; i++) {
    uint64_t out0, out1, out2, out3, out4, out5, out6, out7;
    uint64_t tmp0, tmp1, tmp2;
    uint64_t in0 = out_ptr[0];
    uint64_t in1 = out_ptr[width];
    uint64_t in2 = out_ptr[width * 2];
    uint64_t in3 = out_ptr[width * 3];
    uint64_t in4 = out_ptr[width * 4];
    uint64_t in5 = out_ptr[width * 5];
    uint64_t in6 = out_ptr[width * 6];
    uint64_t in7 = out_ptr[width * 7];
    tmp0 = in4 ^ in7;
    tmp1 = in5 ^ in7;
    out7 = in0 ^ in1 ^ in6;
    out5 = tmp0 ^ in6;
    out2 = tmp0 ^ in3;
    out6 = tmp1 ^ in0;
    out4 = tmp1 ^ in3 ^ in6;
    tmp2 = out5 ^ out7 ^ in2;
    out1 = tmp2 ^ out2;
    out0 = tmp2 ^ in4;
    out3 = tmp2 ^ in5;
    out_ptr[0] = out0 ^ in_ptr[0];
    out_ptr[width] = out1 ^ in_ptr[width];
    out_ptr[width * 2] = out2 ^ in_ptr[width * 2];
    out_ptr[width * 3] = out3 ^ in_ptr[width * 3];
    out_ptr[width * 4] = out4 ^ in_ptr[width * 4];
    out_ptr[width * 5] = out5 ^ in_ptr[width * 5];
    out_ptr[width * 6] = out6 ^ in_ptr[width * 6];
    out_ptr[width * 7] = out7 ^ in_ptr[width * 7];
    in_ptr++;
    out_ptr++;
  }
}
```