user: 下面的汇编指令对应的C源代码是什么:

```asm
<ezplib_get_subrule>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  sub    $0x150,%rsp
  mov    %rdi,-0x2138(%rbp)
  mov    %esi,-0x213c(%rbp)
  mov    %edx,-0x2140(%rbp)
  mov    %ecx,-0x2144(%rbp)
  mov    %r8,-0x2150(%rbp)
  mov    %r9d,-0x2148(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  cmpq   $0x0,-0x2138(%rbp)
  je     L127
  mov    -0x2138(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L127
  cmpq   $0x0,-0x2150(%rbp)
  jne    L86
L127:
  mov    $0xffffffff,%eax
  jmp    L82
L86:
  mov    -0x2140(%rbp),%eax
  cmp    -0x2144(%rbp),%eax
  jle    L36
  mov    $0xffffffff,%eax
  jmp    L82
L36:
  mov    -0x2148(%rbp),%edx
  mov    -0x2150(%rbp),%rax
  mov    $0x0,%esi
  mov    %rax,%rdi
  call   <memset@plt>
  lea    -0x2010(%rbp),%rdx
  mov    -0x213c(%rbp),%esi
  mov    -0x2138(%rbp),%rax
  mov    $0x2000,%ecx
  mov    %rax,%rdi
  call   <ezplib_get_rule@plt>
  mov    %eax,-0x2128(%rbp)
  cmpl   $0x0,-0x2128(%rbp)
  jns    L71
  mov    -0x2128(%rbp),%eax
  jmp    L82
L71:
  lea    -0x2120(%rbp),%rdx
  lea    -0x2010(%rbp),%rax
  mov    $0x22,%ecx
  lea    D0(%rip),%rsi
  mov    %rax,%rdi
  call   <sep_string@plt>
  movl   $0x0,-0x212c(%rbp)
  mov    -0x212c(%rbp),%eax
  mov    %eax,-0x2130(%rbp)
  jmp    L14
L67:
  mov    -0x2130(%rbp),%eax
  cltq
  mov    -0x2120(%rbp,%rax,8),%rax
  test   %rax,%rax
  je     L44
  mov    -0x2140(%rbp),%eax
  cmp    -0x2130(%rbp),%eax
  jg     L166
  mov    -0x2130(%rbp),%eax
  cmp    -0x2144(%rbp),%eax
  jg     L166
  mov    -0x2130(%rbp),%eax
  cltq
  mov    -0x2120(%rbp,%rax,8),%rdx
  mov    -0x2148(%rbp),%eax
  sub    -0x212c(%rbp),%eax
  mov    -0x212c(%rbp),%ecx
  movslq %ecx,%rsi
  mov    -0x2150(%rbp),%rcx
  lea    (%rsi,%rcx,1),%rdi
  lea    D0(%rip),%r8
  mov    %rdx,%rcx
  lea    D12(%rip),%rdx
  mov    %eax,%esi
  call   <snprintf@plt>
  mov    %eax,-0x2124(%rbp)
  mov    -0x2124(%rbp),%eax
  add    %eax,-0x212c(%rbp)
L166:
  addl   $0x1,-0x2130(%rbp)
L14:
  cmpl   $0x21,-0x2130(%rbp)
  jle    L67
  jmp    L144
L44:
  nop
L144:
  cmpl   $0x0,-0x212c(%rbp)
  jle    L99
  mov    -0x212c(%rbp),%eax
  cltq
  lea    -0x1(%rax),%rdx
  mov    -0x2150(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L99
  mov    -0x212c(%rbp),%eax
  cltq
  lea    -0x1(%rax),%rdx
  mov    -0x2150(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
L99:
  mov    $0x0,%eax
L82:
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L155
  call   <__stack_chk_fail@plt>
L155:
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_tag: D0
type_guessed: string
name: unpack_data
arguments: data_tag: D12
type_guessed: string

tool: D0: ^

tool: D12: %s%s

assistant: ```c
int ezplib_get_subrule(const char *rule_set, int nth, int start, int end, char *buf, int bsize) {
  char *ptr_array[34];
  char word[8192];
  int ret, i, j, len;
  if (!rule_set || !*rule_set || !buf) {
    return -1;
  }
  if (start > end) {
    return -1;
  }
  memset(buf, '\0', bsize);
  ret = ezplib_get_rule(rule_set, nth, word, 8192);
  if (ret < 0) {
    return ret;
  }
  sep_string(word, "^", ptr_array, 34);
  for (i = j = 0; i < 34; i++) {
    if (!ptr_array[i]) {
      break;
    }
    if (start <= i && i <= end) {
      len = snprintf(buf + j, bsize - j, "%s%s", ptr_array[i], "^");
      j += len;
    }
  }
  if (j > 0 && buf[j - 1]) {
    buf[j - 1] = '\0';
  }
  return 0;
}
```