user: 这段汇编代码用C语言表示是什么样的：

```asm
<DoMatch>:
  endbr64
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  movzbl (%rsi),%eax
  mov    %rsi,%rbx
  test   %al,%al
  je     L217
  movzbl 0x0(%rbp),%edx
  cmp    $0x2a,%al
  je     L71
L332:
  test   %dl,%dl
  je     L93
L71:
  cmp    $0x5b,%al
  je     L338
L144:
  jg     L138
  cmp    $0x2a,%al
  je     L233
  cmp    $0x3f,%al
  jne    L0
L270:
  movzbl 0x1(%rbx),%eax
  add    $0x1,%rbp
  lea    0x1(%rbx),%rdx
  test   %al,%al
  je     L217
  mov    %rdx,%rbx
  movzbl 0x0(%rbp),%edx
  cmp    $0x2a,%al
  jne    L332
  cmp    $0x5b,%al
  jne    L144
L338:
  movzbl 0x1(%rbx),%esi
  xor    %r9d,%r9d
  cmp    $0x5e,%sil
  sete   %r9b
  jne    L298
  movzbl 0x2(%rbx),%esi
  add    $0x1,%rbx
L298:
  cmp    $0x5d,%sil
  je     L300
  cmp    $0x2d,%sil
  je     L300
  movsbl (%rbx),%edi
  mov    %esi,%eax
  mov    %rbx,%rcx
  xor    %r8d,%r8d
L77:
  lea    0x1(%rcx),%rbx
  test   %al,%al
  je     L266
  cmp    $0x5d,%al
  je     L266
  movsbl %dl,%r10d
  jmp    L75
L234:
  movzbl (%rbx),%esi
  cmp    %al,%dl
  jne    L285
L205:
  mov    %rbx,%rcx
  mov    $0x1,%r8d
  movsbl %sil,%edi
  add    $0x1,%rbx
  movzbl 0x1(%rcx),%eax
  test   %al,%al
  je     L266
L302:
  cmp    $0x5d,%al
  je     L266
L75:
  cmp    $0x2d,%al
  jne    L234
  movzbl 0x2(%rcx),%esi
  cmp    $0x5d,%sil
  je     L234
  lea    0x2(%rcx),%rbx
  cmp    %sil,%dl
  jg     L285
  cmp    %edi,%r10d
  jge    L205
L285:
  mov    %rbx,%rcx
  movsbl %sil,%edi
  add    $0x1,%rbx
  movzbl 0x1(%rcx),%eax
  test   %al,%al
  jne    L302
L266:
  cmp    %r8d,%r9d
  jne    L270
L4:
  xor    %eax,%eax
L150:
  pop    %rbx
  pop    %rbp
  pop    %r12
  ret
  nopl   0x0(%rax,%rax,1)
L138:
  cmp    $0x5c,%al
  sete   %al
  movzbl %al,%eax
  add    %rax,%rbx
L0:
  cmp    (%rbx),%dl
  jne    L4
  jmp    L270
  nopl   0x0(%rax)
L233:
  movzbl 0x1(%rbx),%ecx
  add    $0x1,%rbx
  cmp    $0x2a,%cl
  je     L233
  mov    $0x1,%eax
  test   %cl,%cl
  je     L150
  test   %dl,%dl
  je     L93
L322:
  mov    %rbx,%rsi
  mov    %rbp,%rdi
  lea    0x1(%rbp),%r12
  call   <DoMatch>
  test   %eax,%eax
  jne    L150
  movzbl 0x1(%rbp),%edx
  mov    %r12,%rbp
  test   %dl,%dl
  jne    L322
L93:
  pop    %rbx
  mov    $0xffffffff,%eax
  pop    %rbp
  pop    %r12
  ret
  nopw   0x0(%rax,%rax,1)
L300:
  xor    %r8d,%r8d
  cmp    %sil,%dl
  movzbl 0x2(%rbx),%eax
  lea    0x1(%rbx),%rcx
  sete   %r8b
  movsbl %sil,%edi
  jmp    L77
  nopl   0x0(%rax,%rax,1)
L217:
  xor    %eax,%eax
  cmpb   $0x0,0x0(%rbp)
  sete   %al
  jmp    L150
```

assistant: ```c
int DoMatch(register const char *text, register const char *p) {
  register int last;
  register int matched;
  register int reverse;
  for (; *p; text++, p++) {
    if (*text == '\0' && *p != '*')
      return -1;
    switch (*p) {
    case '\\':
      p++;
    default:
      if (*text != *p)
        return 0;
      continue;
    case '?':
      continue;
    case '*':
      while (*++p == '*')
        continue;
      if (*p == '\0')
        return 1;
      while (*text)
        if ((matched = DoMatch(text++, p)) != 0)
          return matched;
      return -1;
    case '[':
      reverse = p[1] == '^' ? 1 : 0;
      if (reverse)
        p++;
      matched = 0;
      if (p[1] == ']' || p[1] == '-')
        if (*++p == *text)
          matched = 1;
      for (last = *p; *++p && *p != ']'; last = *p)
        if (*p == '-' && p[1] != ']' ? *text <= *++p && *text >= last : *text == *p)
          matched = 1;
      if (matched == reverse)
        return 0;
      continue;
    }
  }
  return *text == '\0';
}
```