user: 请显示与下面列出的汇编代码相匹配的C语言实现：

```asm
<break_line>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x28,%rsp
  mov    %rsi,0x18(%rsp)
  test   %rdi,%rdi
  je     L128
  movzbl (%rdi),%edx
  mov    %rdi,%r12
  test   %dl,%dl
  je     L35
  movl   $0x0,0x14(%rsp)
  xor    %r13d,%r13d
  xor    %r14d,%r14d
  nopl   0x0(%rax)
L271:
  cmp    $0x22,%dl
  je     L13
  xor    %esi,%esi
  cmp    $0x27,%dl
  je     L13
L234:
  mov    %r12,%rbp
  mov    %edx,%eax
  jmp    L227
  nopl   0x0(%rax,%rax,1)
L238:
  cmp    %sil,%al
  je     L36
  test   %sil,%sil
  jne    L222
L36:
  cmp    $0x2c,%al
  je     L255
L222:
  add    $0x1,%rbp
L2:
  movzbl 0x0(%rbp),%eax
  test   %al,%al
  je     L255
L227:
  cmp    $0x5c,%al
  jne    L238
  add    $0x2,%rbp
  jmp    L2
  nopl   0x0(%rax)
L17:
  movzbl 0x1(%r12),%edx
  add    $0x1,%r12
  test   %dl,%dl
  je     L70
L255:
  cmp    $0x20,%dl
  je     L17
  cmp    $0x9,%dl
  je     L17
L70:
  mov    %ebp,%r15d
  sub    %r12d,%r15d
  test   %r15d,%r15d
  je     L215
  movslq %r15d,%rbx
  jmp    L239
  nopl   0x0(%rax)
L262:
  sub    $0x1,%rbx
  mov    %ebx,%r15d
  test   %ebx,%ebx
  je     L215
L239:
  movzbl -0x1(%r12,%rbx,1),%eax
  mov    %ebx,%r15d
  cmp    $0x20,%al
  je     L262
  cmp    $0x9,%al
  je     L262
  lea    0x1(%rbx),%edi
L258:
  call   <malloc@plt>
  mov    %r15d,%edx
  mov    %r12,%rsi
  mov    %rax,%rdi
  mov    %rax,0x8(%rsp)
  call   <strncpy@plt>
  mov    0x8(%rsp),%r9
  mov    %r14,%rdi
  lea    0x10(%r13),%esi
  addl   $0x1,0x14(%rsp)
  movb   $0x0,(%r9,%rbx,1)
  call   <realloc@plt>
  mov    0x8(%rsp),%r9
  mov    %rax,%r14
  lea    0x8(%r13),%rax
  mov    %r9,(%r14,%r13,1)
  movq   $0x0,0x8(%r14,%r13,1)
  cmpb   $0x0,0x0(%rbp)
  jne    L254
L178:
  mov    0x18(%rsp),%rax
  mov    0x14(%rsp),%ecx
  mov    %ecx,(%rax)
  add    $0x28,%rsp
  mov    %r14,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
  nopw   0x0(%rax,%rax,1)
L13:
  movzbl 0x1(%r12),%eax
  lea    0x1(%r12),%rbp
  test   %al,%al
  je     L24
  mov    %edx,%esi
  mov    %rbp,%r12
  mov    %eax,%edx
  jmp    L234
  nopl   0x0(%rax,%rax,1)
L254:
  movzbl 0x1(%rbp),%edx
  lea    0x1(%rbp),%r12
  test   %dl,%dl
  je     L178
  mov    %rax,%r13
  jmp    L271
  nopl   0x0(%rax)
L215:
  xor    %ebx,%ebx
  mov    $0x1,%edi
  jmp    L258
L24:
  mov    %rbp,%r12
  xor    %ebx,%ebx
  mov    $0x1,%edi
  xor    %r15d,%r15d
  jmp    L258
L35:
  movl   $0x0,0x14(%rsp)
  xor    %r14d,%r14d
  jmp    L178
L128:
  movl   $0x0,0x14(%rsp)
  xor    %r14d,%r14d
  jmp    L178
```

assistant: ```c
char **break_line(const char *b, int *n) {
  char **ret = NULL;
  char *t = NULL, *s;
  char term;
  int n_strs = 0, len;
  while (b && *b) {
    term = 0;
    if (*b == '"' || *b == '\'')
      term = *b++;
    t = (char *)b;
    while (*b) {
      if (*b == '\\')
        b += 2;
      else if (term && *b != term)
        b++;
      else if (*b != ',')
        b++;
      else
        break;
    }
    while (*t && (*t == ' ' || *t == '\t'))
      t++;
    len = b - t;
    if (term && *b == term)
      len--;
    while (len && (t[len - 1] == ' ' || t[len - 1] == '\t'))
      len--;
    s = malloc(len + 1);
    strncpy(s, t, len);
    s[len] = 0;
    n_strs++;
    ret = realloc(ret, (n_strs + 1) * sizeof(char *));
    ret[n_strs - 1] = s;
    ret[n_strs] = NULL;
    if (*b)
      b++;
  }
  *n = n_strs;
  return ret;
}
```