user: 这段汇编代码用C语言表示是什么样的:

```asm
<padMsg>:
  endbr64
  mov    %esi,%eax
  mov    %esi,%r11d
  mov    %rdi,%rcx
  push   %rbx
  and    $0xffffffc0,%eax
  and    $0x3f,%r11d
  mov    %rdx,%rdi
  cltq
  lea    -0x1(%r11),%r10d
  shl    $0x3,%rax
  lea    (%rcx,%rax,1),%r9
  cmp    $0x37,%r11d
  jg     L55
  mov    $0x37,%edx
  mov    $0x1,%r8d
  sub    %r11d,%edx
  test   %r11d,%r11d
  je     L35
L52:
  lea    0x8(%rcx,%rax,1),%rax
  cmp    %rax,%rdi
  je     L36
  cmp    $0x2,%r10d
  jbe    L36
  mov    %r11d,%ecx
  xor    %eax,%eax
  shr    %ecx
  shl    $0x4,%rcx
  nopl   (%rax)
L94:
  movdqu (%r9,%rax,1),%xmm3
  movups %xmm3,(%rdi,%rax,1)
  add    $0x10,%rax
  cmp    %rax,%rcx
  jne    L94
  mov    %r11d,%ebx
  and    $0xfffffffe,%ebx
  mov    %ebx,%eax
  shl    $0x3,%rax
  lea    (%rdi,%rax,1),%rcx
  add    %r9,%rax
  cmp    %ebx,%r11d
  je     L87
  mov    (%rax),%rax
  mov    %rax,(%rcx)
L87:
  mov    %r10d,%r10d
  lea    0x8(%rdi,%r10,8),%rdi
L35:
  movq   $0x80,(%rdi)
  lea    0x8(%rdi),%r9
  test   %edx,%edx
  je     L70
  movq   $0x0,0x8(%rdi)
  add    $0x10,%rdi
  mov    %r9,%rcx
  movslq %edx,%rdx
  and    $0xfffffffffffffff8,%rdi
  shl    $0x3,%rdx
  sub    %rdi,%rcx
  mov    %edx,%eax
  movq   $0x0,-0x8(%r9,%rax,1)
  add    %edx,%ecx
  xor    %eax,%eax
  add    %rdx,%r9
  shr    $0x3,%ecx
  rep stos %rax,%es:(%rdi)
L70:
  lea    0x0(,%rsi,8),%eax
  mov    $0x38,%ecx
  mov    %eax,%ebx
  mov    %eax,%edx
  mov    %eax,%esi
  sar    %cl,%ebx
  mov    $0x30,%ecx
  sar    %cl,%edx
  movd   %ebx,%xmm0
  mov    $0x28,%ecx
  mov    %eax,%ebx
  sar    %cl,%ebx
  mov    $0x20,%ecx
  movd   %edx,%xmm5
  mov    %eax,%edx
  sar    %cl,%esi
  movd   %ebx,%xmm1
  punpckldq %xmm5,%xmm0
  shr    $0x18,%edx
  movd   %esi,%xmm4
  punpckldq %xmm4,%xmm1
  punpcklqdq %xmm1,%xmm0
  pand   D4(%rip),%xmm0
  pxor   %xmm1,%xmm1
  pcmpgtd %xmm0,%xmm1
  movdqa %xmm0,%xmm2
  punpckhdq %xmm1,%xmm0
  punpckldq %xmm1,%xmm2
  movups %xmm0,0x10(%r9)
  movd   %edx,%xmm0
  mov    %eax,%edx
  sar    $0x10,%edx
  movups %xmm2,(%r9)
  movzbl %dl,%edx
  movq   %rdx,%xmm6
  movzbl %ah,%edx
  and    $0xf8,%eax
  punpcklqdq %xmm6,%xmm0
  movslq %edx,%rsi
  movq   %rax,%xmm7
  mov    %r8d,%eax
  movups %xmm0,0x20(%r9)
  movq   %rsi,%xmm0
  punpcklqdq %xmm7,%xmm0
  movups %xmm0,0x30(%r9)
  pop    %rbx
  ret
  nopl   0x0(%rax,%rax,1)
L55:
  mov    $0x77,%edx
  mov    $0x2,%r8d
  sub    %r11d,%edx
  jmp    L52
  nopl   0x0(%rax,%rax,1)
L36:
  mov    %r10d,%r11d
  xor    %eax,%eax
  nopl   (%rax)
L89:
  mov    (%r9,%rax,8),%rcx
  mov    %rcx,(%rdi,%rax,8)
  mov    %rax,%rcx
  add    $0x1,%rax
  cmp    %r11,%rcx
  jne    L89
  jmp    L87
```

tool_calls: name: read_data
arguments: data_label: D4
type_infered: byte[16]

tool: {"D4": [0xff,0x00,0x00,0x00,0xff,0x00,0x00,0x00,0xff,0x00,0x00,0x00,0xff,0x00,0x00,0x00]}

assistant: ```c
int padMsg(const void *msg, uint64_t bytes, uint8_t *lastBlock) {
  int msgBytesInLast = (int)(bytes & 63);
  int extraBlocks = (msgBytesInLast + 9) > 64 ? 2 : 1;
  int numZeroBytes = extraBlocks * 64 - 9 - msgBytesInLast;
  const uint8_t *msgLast = (uint8_t *)msg + (bytes & ~63);
  while (msgBytesInLast--)
    *lastBlock++ = *msgLast++;
  *lastBlock++ = 0x80;
  while (numZeroBytes--)
    *lastBlock++ = 0;
  bytes *= 8;
  *lastBlock++ = (uint8_t)(bytes >> 56 & 0xff);
  *lastBlock++ = (uint8_t)(bytes >> 48 & 0xff);
  *lastBlock++ = (uint8_t)(bytes >> 40 & 0xff);
  *lastBlock++ = (uint8_t)(bytes >> 32 & 0xff);
  *lastBlock++ = (uint8_t)(bytes >> 24 & 0xff);
  *lastBlock++ = (uint8_t)(bytes >> 16 & 0xff);
  *lastBlock++ = (uint8_t)(bytes >> 8 & 0xff);
  *lastBlock++ = (uint8_t)(bytes >> 0 & 0xff);
  return extraBlocks;
}
```