user: 请从给定的汇编指令推导出C源代码：

```asm
<taa_h264_inverse_hadamard_4x4>:
  endbr64
  mov    %rsi,%rdx
  mov    %rdi,%rcx
  mov    %rsi,%rax
  sub    %rdi,%rdx
  sub    $0x4,%rdx
  cmp    $0x38,%rdx
  jbe    L5
  movdqu (%rdi),%xmm1
  movdqu 0x10(%rdi),%xmm4
  movdqu 0x20(%rdi),%xmm3
  movdqu 0x30(%rdi),%xmm0
  movdqa %xmm1,%xmm2
  shufps $0xdd,%xmm4,%xmm1
  shufps $0x88,%xmm4,%xmm2
  movdqa %xmm3,%xmm4
  shufps $0xdd,%xmm0,%xmm3
  shufps $0x88,%xmm0,%xmm4
  movdqa %xmm2,%xmm0
  shufps $0xdd,%xmm4,%xmm2
  shufps $0x88,%xmm4,%xmm0
  movdqa %xmm1,%xmm4
  shufps $0xdd,%xmm3,%xmm1
  shufps $0x88,%xmm3,%xmm4
  movdqa %xmm0,%xmm3
  movdqa %xmm4,%xmm5
  psubd  %xmm2,%xmm0
  paddd  %xmm2,%xmm3
  psubd  %xmm1,%xmm5
  movdqa %xmm0,%xmm2
  paddd  %xmm1,%xmm4
  movdqa %xmm3,%xmm1
  psubd  %xmm5,%xmm0
  paddd  %xmm4,%xmm1
  psubd  %xmm4,%xmm3
  paddd  %xmm5,%xmm2
  movdqa %xmm1,%xmm4
  punpckhdq %xmm0,%xmm1
  punpckldq %xmm0,%xmm4
  movdqa %xmm2,%xmm5
  movdqa %xmm2,%xmm0
  punpckldq %xmm3,%xmm5
  movdqa %xmm4,%xmm2
  punpckhdq %xmm3,%xmm0
  punpckldq %xmm5,%xmm2
  punpckhdq %xmm5,%xmm4
  movups %xmm2,(%rsi)
  movdqa %xmm1,%xmm2
  punpckhdq %xmm0,%xmm1
  movdqa %xmm4,%xmm3
  punpckldq %xmm0,%xmm2
  movups %xmm4,0x10(%rsi)
  movups %xmm2,0x20(%rsi)
  movdqa %xmm2,%xmm7
  movups %xmm1,0x30(%rsi)
L8:
  movdqu (%rax),%xmm6
  paddd  %xmm6,%xmm2
  movdqa %xmm6,%xmm0
  movdqu 0x30(%rax),%xmm6
  psubd  %xmm7,%xmm0
  movdqu 0x10(%rax),%xmm7
  movdqa %xmm6,%xmm1
  psubd  %xmm6,%xmm3
  paddd  %xmm7,%xmm1
  movdqa %xmm1,%xmm4
  paddd  %xmm2,%xmm4
  psubd  %xmm1,%xmm2
  movups %xmm4,(%rax)
  movdqa %xmm3,%xmm4
  paddd  %xmm0,%xmm4
  psubd  %xmm3,%xmm0
  movups %xmm2,0x30(%rax)
  movups %xmm4,0x10(%rax)
  movups %xmm0,0x20(%rax)
  ret
  nopw   0x0(%rax,%rax,1)
L5:
  mov    (%rdi),%edx
  mov    0x4(%rcx),%esi
  mov    0x8(%rdi),%edi
  mov    %esi,%r8d
  add    0xc(%rcx),%esi
  sub    0xc(%rcx),%r8d
  add    %edx,%edi
  sub    0x8(%rcx),%edx
  lea    (%rsi,%rdi,1),%r9d
  sub    %esi,%edi
  mov    %r9d,(%rax)
  lea    (%rdx,%r8,1),%r9d
  sub    %r8d,%edx
  mov    %r9d,0x4(%rax)
  mov    %edx,0x8(%rax)
  mov    %edi,0xc(%rax)
  mov    0x10(%rcx),%edx
  mov    0x14(%rcx),%esi
  mov    0x18(%rcx),%edi
  mov    %esi,%r8d
  add    %edx,%edi
  add    0x1c(%rcx),%esi
  sub    0x1c(%rcx),%r8d
  sub    0x18(%rcx),%edx
  lea    (%rdi,%rsi,1),%r9d
  sub    %esi,%edi
  mov    %r9d,0x10(%rax)
  lea    (%rdx,%r8,1),%r9d
  sub    %r8d,%edx
  mov    %r9d,0x14(%rax)
  mov    %edx,0x18(%rax)
  mov    %edi,0x1c(%rax)
  mov    0x20(%rcx),%edx
  mov    0x24(%rcx),%esi
  mov    0x28(%rcx),%edi
  mov    %esi,%r8d
  add    %edx,%edi
  add    0x2c(%rcx),%esi
  sub    0x2c(%rcx),%r8d
  sub    0x28(%rcx),%edx
  lea    (%rdi,%rsi,1),%r9d
  sub    %esi,%edi
  mov    %r9d,0x20(%rax)
  lea    (%rdx,%r8,1),%r9d
  sub    %r8d,%edx
  mov    %edi,0x2c(%rax)
  mov    %r9d,0x24(%rax)
  mov    %edx,0x28(%rax)
  mov    0x30(%rcx),%edx
  mov    0x38(%rcx),%esi
  add    %edx,%esi
  sub    0x38(%rcx),%edx
  mov    0x34(%rcx),%r8d
  mov    0x3c(%rcx),%ecx
  movdqu 0x20(%rax),%xmm2
  mov    %r8d,%edi
  movdqu 0x20(%rax),%xmm7
  movdqu 0x10(%rax),%xmm3
  sub    %ecx,%edi
  add    %r8d,%ecx
  lea    (%rcx,%rsi,1),%r8d
  sub    %ecx,%esi
  mov    %r8d,0x30(%rax)
  lea    (%rdi,%rdx,1),%r8d
  sub    %edi,%edx
  mov    %r8d,0x34(%rax)
  mov    %edx,0x38(%rax)
  mov    %esi,0x3c(%rax)
  jmp    L8
```

assistant: ```c
void taa_h264_inverse_hadamard_4x4(int16_t *coeff, int16_t *block) {
  int tmp0, tmp1, tmp2, tmp3;
  for (int i = 0; i < 4; i++) {
    tmp0 = coeff[i * 4 + 0] + coeff[i * 4 + 2];
    tmp1 = coeff[i * 4 + 0] - coeff[i * 4 + 2];
    tmp2 = coeff[i * 4 + 1] - coeff[i * 4 + 3];
    tmp3 = coeff[i * 4 + 1] + coeff[i * 4 + 3];
    block[i * 4 + 0] = (int16_t)(tmp0 + tmp3);
    block[i * 4 + 1] = (int16_t)(tmp1 + tmp2);
    block[i * 4 + 2] = (int16_t)(tmp1 - tmp2);
    block[i * 4 + 3] = (int16_t)(tmp0 - tmp3);
  }
  for (int i = 0; i < 4; i++) {
    tmp0 = block[0 * 4 + i] + block[2 * 4 + i];
    tmp1 = block[0 * 4 + i] - block[2 * 4 + i];
    tmp2 = block[1 * 4 + i] - block[3 * 4 + i];
    tmp3 = block[1 * 4 + i] + block[3 * 4 + i];
    block[0 * 4 + i] = (int16_t)(tmp0 + tmp3);
    block[1 * 4 + i] = (int16_t)(tmp1 + tmp2);
    block[2 * 4 + i] = (int16_t)(tmp1 - tmp2);
    block[3 * 4 + i] = (int16_t)(tmp0 - tmp3);
  }
}
```