user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<escape>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %edx,-0x34(%rbp)
  mov    -0x30(%rbp),%rax
  mov    %rax,-0x8(%rbp)
  mov    -0x28(%rbp),%rax
  mov    $0xa,%esi
  mov    %rax,%rdi
  call   <strchr@plt>
  test   %rax,%rax
  setne  %al
  movzbl %al,%eax
  mov    %eax,-0x10(%rbp)
  movl   $0x0,-0x14(%rbp)
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0xc(%rbp)
  cmpl   $0x0,-0xc(%rbp)
  jle    L44
  mov    -0xc(%rbp),%eax
  cltq
  lea    -0x1(%rax),%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0xa,%al
  jne    L44
  movl   $0x1,-0x14(%rbp)
L44:
  mov    -0x8(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x8(%rbp)
  movb   $0x22,(%rax)
  subl   $0x1,-0x34(%rbp)
  cmpl   $0x0,-0x10(%rbp)
  je     L113
  mov    -0x8(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x8(%rbp)
  movb   $0x22,(%rax)
  mov    -0x8(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x8(%rbp)
  movb   $0xa,(%rax)
  mov    -0x8(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x8(%rbp)
  movb   $0x22,(%rax)
  subl   $0x3,-0x34(%rbp)
  jmp    L113
L65:
  mov    -0x28(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0x22,%al
  jne    L24
  mov    -0x8(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x8(%rbp)
  movb   $0x5c,(%rax)
  jmp    L68
L24:
  mov    -0x28(%rbp),%rax
  movzbl (%rax),%eax
  cmp    $0xa,%al
  jne    L68
  mov    -0x8(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x8(%rbp)
  movb   $0x5c,(%rax)
  mov    -0x8(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x8(%rbp)
  movb   $0x6e,(%rax)
  mov    -0x8(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x8(%rbp)
  movb   $0x22,(%rax)
  mov    -0x8(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x8(%rbp)
  movb   $0xa,(%rax)
  mov    -0x8(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x8(%rbp)
  movb   $0x22,(%rax)
  subl   $0x5,-0x34(%rbp)
  addq   $0x1,-0x28(%rbp)
  jmp    L51
L68:
  mov    -0x28(%rbp),%rdx
  lea    0x1(%rdx),%rax
  mov    %rax,-0x28(%rbp)
  mov    -0x8(%rbp),%rax
  lea    0x1(%rax),%rcx
  mov    %rcx,-0x8(%rbp)
  movzbl (%rdx),%edx
  mov    %dl,(%rax)
L51:
  subl   $0x1,-0x34(%rbp)
L113:
  mov    -0x28(%rbp),%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L15
  cmpl   $0x1,-0x34(%rbp)
  jg     L65
L15:
  cmpl   $0x0,-0x10(%rbp)
  je     L90
  cmpl   $0x0,-0x14(%rbp)
  je     L90
  subq   $0x3,-0x8(%rbp)
L90:
  mov    -0x8(%rbp),%rax
  lea    0x1(%rax),%rdx
  mov    %rdx,-0x8(%rbp)
  movb   $0x22,(%rax)
  mov    -0x8(%rbp),%rax
  movb   $0x0,(%rax)
  mov    -0x30(%rbp),%rax
  leave
  ret
```

assistant: ```c
char *escape(const char *text, char *bf, int len) {
  char *bfp = bf;
  int multiline = strchr(text, '\n') != NULL;
  int eol = 0;
  int textlen = strlen(text);
  if ((textlen > 0) && (text[textlen - 1] == '\n'))
    eol = 1;
  *bfp++ = '"';
  --len;
  if (multiline) {
    *bfp++ = '"';
    *bfp++ = '\n';
    *bfp++ = '"';
    len -= 3;
  }
  while (*text != '\0' && len > 1) {
    if (*text == '"')
      *bfp++ = '\\';
    else if (*text == '\n') {
      *bfp++ = '\\';
      *bfp++ = 'n';
      *bfp++ = '"';
      *bfp++ = '\n';
      *bfp++ = '"';
      len -= 5;
      ++text;
      goto next;
    }
    *bfp++ = *text++;
  next:
    --len;
  }
  if (multiline && eol)
    bfp -= 3;
  *bfp++ = '"';
  *bfp = '\0';
  return bf;
}
```