user: 这段汇编代码用C语言表示是什么样的?

```asm
<llib_led_regex_brack>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  mov    %ecx,%ebx
  sub    $0x38,%rsp
  mov    (%rsi),%rcx
  mov    (%rdx),%r15d
  mov    %rsi,(%rsp)
  mov    %fs:0x28,%rax
  mov    %rax,0x28(%rsp)
  xor    %eax,%eax
  mov    %rcx,0x20(%rsp)
  mov    %r15d,0x1c(%rsp)
  test   %r15d,%r15d
  je     L212
  movzbl (%rcx),%eax
  mov    %edi,%ebp
  mov    %rdx,%r14
  cmp    $0x5e,%al
  je     L38
  movl   $0x0,0xc(%rsp)
  cmp    $0x2a,%al
  je     L38
L55:
  cmp    $0x5d,%al
  je     L75
  mov    0x1c(%rsp),%edx
  mov    %rcx,%rdi
  xor    %r15d,%r15d
L57:
  lea    0x1c(%rsp),%r12
L172:
  cmp    $0x5d,%al
  je     L151
L210:
  lea    0x20(%rsp),%r13
  mov    %r12,%rdx
  mov    %ebp,%edi
  mov    %r13,%rsi
  call   <llib_utf8_escape@plt>
  mov    0x1c(%rsp),%edx
  mov    0x20(%rsp),%rdi
  mov    %eax,%ecx
  cmp    $0x1,%edx
  jle    L72
  movzbl (%rdi),%eax
  cmp    $0x2d,%al
  je     L106
L154:
  cmp    %ecx,%ebx
  je     L90
L200:
  movzbl (%rdi),%eax
  cmp    $0x5d,%al
  jne    L210
L151:
  add    $0x1,%rdi
  sub    $0x1,%edx
L186:
  mov    (%rsp),%rax
  mov    %rdi,(%rax)
  mov    0xc(%rsp),%eax
  mov    %edx,(%r14)
  test   %eax,%eax
  jne    L221
L212:
  mov    0x28(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L178
  add    $0x38,%rsp
  mov    %r15d,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopl   0x0(%rax,%rax,1)
L38:
  lea    0x1(%rcx),%rdx
  sub    $0x1,%r15d
  mov    %rdx,0x20(%rsp)
  mov    %r15d,0x1c(%rsp)
  jne    L83
  mov    (%rsp),%rax
  mov    %rdx,(%rax)
  movl   $0x0,(%r14)
L221:
  xor    $0x1,%r15d
  jmp    L212
  nopl   0x0(%rax)
L72:
  cmp    %eax,%ebx
  mov    $0x1,%eax
  cmove  %eax,%r15d
L176:
  test   %edx,%edx
  jne    L200
  jmp    L186
  nopl   0x0(%rax,%rax,1)
L106:
  cmpb   $0x5d,0x1(%rdi)
  je     L154
  sub    $0x1,%edx
  add    $0x1,%rdi
  mov    %r13,%rsi
  mov    %ecx,0x8(%rsp)
  mov    %edx,0x1c(%rsp)
  mov    %r12,%rdx
  mov    %rdi,0x20(%rsp)
  mov    %ebp,%edi
  call   <llib_utf8_escape@plt>
  mov    0x8(%rsp),%ecx
  mov    0x1c(%rsp),%edx
  mov    0x20(%rsp),%rdi
  cmp    %ecx,%ebx
  jl     L176
  cmp    %eax,%ebx
  jg     L176
  mov    $0x1,%r15d
  jmp    L176
  nopl   0x0(%rax,%rax,1)
L90:
  mov    $0x1,%r15d
  jmp    L172
  nopl   0x0(%rax,%rax,1)
L75:
  lea    -0x1(%r15),%edx
  lea    0x1(%rcx),%rdi
  xor    %r15d,%r15d
  cmp    $0x5d,%ebx
  mov    %rdi,0x20(%rsp)
  sete   %r15b
  mov    %edx,0x1c(%rsp)
  test   %edx,%edx
  je     L186
  movzbl 0x1(%rcx),%eax
  jmp    L57
L178:
  call   <__stack_chk_fail@plt>
L83:
  movzbl 0x1(%rcx),%eax
  movl   $0x1,0xc(%rsp)
  mov    %rdx,%rcx
  jmp    L55
```

assistant: ```c
int llib_led_regex_brack(int utf8, unsigned char **a, int *la, int c) {
  int inverse = 0;
  int flag = 0;
  unsigned char *s = *a;
  int l = *la;
  if (!l)
    return 0;
  if (*s == '^' || *s == '*') {
    inverse = 1;
    ++s;
    --l;
  }
  if (l && *s == ']') {
    ++s;
    --l;
    if (c == ']')
      flag = 1;
  }
  while (l) {
    if (*s == ']') {
      ++s;
      --l;
      break;
    } else {
      int cl, cr;
      cl = llib_utf8_escape(utf8, (char **)&s, &l);
      if (l >= 2 && s[0] == '-' && s[1] != ']') {
        --l;
        ++s;
        cr = llib_utf8_escape(utf8, (char **)&s, &l);
        if (c >= cl && c <= cr)
          flag = 1;
      } else if (c == cl)
        flag = 1;
    }
  }
  *a = s;
  *la = l;
  if (inverse)
    return !flag;
  else
    return flag;
}
```