user: 这段汇编代码用C语言表示是什么样的?

```asm
<__md5_Transform>:
  endbr64
  push   %r15
  mov    $0x40,%edx
  xor    %r15d,%r15d
  push   %r14
  lea    D0(%rip),%r14
  push   %r13
  lea    D18(%rip),%r13
  push   %r12
  mov    $0x20,%r12d
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x68,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x58(%rsp)
  xor    %eax,%eax
  lea    0x10(%rsp),%rbp
  mov    %rbp,%rdi
  call   <__md5_Decode@plt>
  mov    0xc(%rbx),%edi
  mov    (%rbx),%eax
  xor    %esi,%esi
  mov    0x4(%rbx),%r9d
  mov    0x8(%rbx),%r10d
  mov    $0xd76aa478,%edx
  lea    D27(%rip),%r8
  mov    %edi,0xc(%rsp)
  mov    %edi,%r11d
  mov    %eax,%edi
  mov    %eax,(%rsp)
  mov    %r9d,0x4(%rsp)
  mov    %r10d,0x8(%rsp)
  jmp    L50
  nopl   0x0(%rax)
L85:
  cmp    $0x1,%eax
  je     L3
  mov    %r10d,%eax
  xor    %r11d,%eax
  and    %r9d,%eax
  xor    %r11d,%eax
  add    %edi,%eax
L56:
  and    $0x3,%ecx
  add    0x10(%rsp,%r15,4),%edx
  add    $0x1,%rsi
  movsbl (%r8,%rcx,1),%edi
  add    %eax,%edx
  mov    %edx,%eax
  mov    %edi,%ecx
  shl    %cl,%eax
  mov    %r12d,%ecx
  sub    %edi,%ecx
  sar    %cl,%edx
  or     %edx,%eax
  add    %r9d,%eax
  cmp    $0x40,%rsi
  je     L65
  movsbq (%r14,%rsi,1),%r15
  mov    0x0(%r13,%rsi,4),%edx
  mov    %r11d,%edi
  mov    %r10d,%r11d
  mov    %r9d,%r10d
  mov    %eax,%r9d
L50:
  lea    0x4(%r8),%rax
  test   $0xf,%sil
  mov    %esi,%ecx
  cmove  %rax,%r8
  mov    %esi,%eax
  sar    $0x4,%eax
  cmp    $0x2,%eax
  je     L47
  cmp    $0x3,%eax
  jne    L85
  mov    %r11d,%eax
  not    %eax
  or     %r9d,%eax
  xor    %r10d,%eax
  add    %edi,%eax
  jmp    L56
  nopl   0x0(%rax)
L47:
  mov    %r9d,%eax
  xor    %r10d,%eax
  xor    %r11d,%eax
  add    %edi,%eax
  jmp    L56
  nopl   (%rax)
L3:
  mov    %r9d,%eax
  xor    %r10d,%eax
  and    %r11d,%eax
  xor    %r10d,%eax
  add    %edi,%eax
  jmp    L56
  nopl   0x0(%rax,%rax,1)
L65:
  mov    %eax,%ecx
  mov    (%rsp),%eax
  xor    %esi,%esi
  mov    $0x40,%edx
  mov    %rbp,%rdi
  add    %r11d,%eax
  mov    %eax,(%rbx)
  mov    0x4(%rsp),%eax
  add    %ecx,%eax
  mov    %eax,0x4(%rbx)
  mov    0x8(%rsp),%eax
  add    %r9d,%eax
  mov    %eax,0x8(%rbx)
  mov    0xc(%rsp),%eax
  add    %r10d,%eax
  mov    %eax,0xc(%rbx)
  call   <memset@plt>
  mov    0x58(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L9
  add    $0x68,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L9:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: data_tag: D0
type_guessed: string

tool: 

assistant: ```c
void __md5_Transform(u_int32_t state[4], const unsigned char block[64]) {
  u_int32_t a, b, c, d, x[16];
  u_int32_t temp;
  const char *ps;
  static const char S[] = {7, 12, 17, 22, 5, 9, 14, 20, 4, 11, 16, 23, 6, 10, 15, 21};
  const u_int32_t *pc;
  const char *pp;
  int i;
  static const u_int32_t C[] = {0xd76aa478, 0xe8c7b756, 0x242070db, 0xc1bdceee, 0xf57c0faf, 0x4787c62a, 0xa8304613, 0xfd469501, 0x698098d8, 0x8b44f7af, 0xffff5bb1, 0x895cd7be, 0x6b901122, 0xfd987193, 0xa679438e, 0x49b40821, 0xf61e2562, 0xc040b340, 0x265e5a51, 0xe9b6c7aa, 0xd62f105d, 0x2441453, 0xd8a1e681, 0xe7d3fbc8, 0x21e1cde6, 0xc33707d6, 0xf4d50d87, 0x455a14ed, 0xa9e3e905, 0xfcefa3f8, 0x676f02d9, 0x8d2a4c8a, 0xfffa3942, 0x8771f681, 0x6d9d6122, 0xfde5380c, 0xa4beea44, 0x4bdecfa9, 0xf6bb4b60, 0xbebfbc70, 0x289b7ec6, 0xeaa127fa, 0xd4ef3085, 0x4881d05, 0xd9d4d039, 0xe6db99e5, 0x1fa27cf8, 0xc4ac5665, 0xf4292244, 0x432aff97, 0xab9423a7, 0xfc93a039, 0x655b59c3, 0x8f0ccc92, 0xffeff47d, 0x85845dd1, 0x6fa87e4f, 0xfe2ce6e0, 0xa3014314, 0x4e0811a1, 0xf7537e82, 0xbd3af235, 0x2ad7d2bb, 0xeb86d391};
  static const char P[] = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 1, 6, 11, 0, 5, 10, 15, 4, 9, 14, 3, 8, 13, 2, 7, 12, 5, 8, 11, 14, 1, 4, 7, 10, 13, 0, 3, 6, 9, 12, 15, 2, 0, 7, 14, 5, 12, 3, 10, 1, 8, 15, 6, 13, 4, 11, 2, 9};
  __md5_Decode(x, block, 64);
  a = state[0];
  b = state[1];
  c = state[2];
  d = state[3];
  pc = C;
  pp = P;
  ps = S - 4;
  for (i = 0; i < 64; i++) {
    if ((i & 0x0f) == 0)
      ps += 4;
    temp = a;
    switch (i >> 4) {
    case 0:
      temp += (((b) & (c)) | ((~b) & (d)));
      break;
    case 1:
      temp += (((b) & (d)) | ((c) & (~d)));
      break;
    case 2:
      temp += ((b) ^ (c) ^ (d));
      break;
    case 3:
      temp += ((c) ^ ((b) | (~d)));
      break;
    }
    temp += x[(int)(*pp++)] + *pc++;
    temp = (((temp) << (ps[i & 3])) | ((temp) >> (32 - (ps[i & 3]))));
    temp += b;
    a = d;
    d = c;
    c = b;
    b = temp;
  }
  state[0] += a;
  state[1] += b;
  state[2] += c;
  state[3] += d;
  memset((void *)x, 0, sizeof(x));
}
```