user: 这段汇编代码用C语言表示是什么样的：

```asm
<abbreviate>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %esi,-0x2c(%rbp)
  cmpq   $0x0,-0x28(%rbp)
  jne    L48
  lea    D0(%rip),%rax
  mov    %rax,%rdi
  call   <my_strdup@plt>
  jmp    L99
L48:
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0x18(%rbp)
  mov    -0x18(%rbp),%eax
  cmp    -0x2c(%rbp),%eax
  jge    L66
  mov    -0x28(%rbp),%rax
  mov    %rax,%rdi
  call   <my_strdup@plt>
  mov    %rax,-0x10(%rbp)
  jmp    L37
L66:
  mov    -0x2c(%rbp),%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x10(%rbp)
  mov    -0x2c(%rbp),%eax
  sub    $0x6,%eax
  mov    %eax,%edx
  shr    $0x1f,%edx
  add    %edx,%eax
  sar    %eax
  mov    %eax,-0x14(%rbp)
  movl   $0x0,-0x1c(%rbp)
  jmp    L69
L23:
  mov    -0x1c(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    -0x1c(%rbp),%edx
  movslq %edx,%rcx
  mov    -0x10(%rbp),%rdx
  add    %rcx,%rdx
  movzbl (%rax),%eax
  mov    %al,(%rdx)
  addl   $0x1,-0x1c(%rbp)
L69:
  mov    -0x1c(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jle    L23
  mov    -0x14(%rbp),%eax
  cltq
  lea    0x1(%rax),%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  movb   $0x20,(%rax)
  mov    -0x14(%rbp),%eax
  cltq
  lea    0x2(%rax),%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  movb   $0x2e,(%rax)
  mov    -0x14(%rbp),%eax
  cltq
  lea    0x3(%rax),%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  movb   $0x2e,(%rax)
  mov    -0x14(%rbp),%eax
  cltq
  lea    0x4(%rax),%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  movb   $0x2e,(%rax)
  mov    -0x14(%rbp),%eax
  cltq
  lea    0x5(%rax),%rdx
  mov    -0x10(%rbp),%rax
  add    %rdx,%rax
  movb   $0x20,(%rax)
  movl   $0x0,-0x1c(%rbp)
  jmp    L89
L39:
  mov    -0x18(%rbp),%eax
  sub    -0x14(%rbp),%eax
  mov    %eax,%edx
  mov    -0x1c(%rbp),%eax
  add    %edx,%eax
  movslq %eax,%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    -0x14(%rbp),%edx
  lea    0x6(%rdx),%ecx
  mov    -0x1c(%rbp),%edx
  add    %ecx,%edx
  movslq %edx,%rcx
  mov    -0x10(%rbp),%rdx
  add    %rcx,%rdx
  movzbl (%rax),%eax
  mov    %al,(%rdx)
  addl   $0x1,-0x1c(%rbp)
L89:
  mov    -0x1c(%rbp),%eax
  cmp    -0x14(%rbp),%eax
  jle    L39
  jmp    L37
L91:
  mov    -0x8(%rbp),%rax
  movb   $0x20,(%rax)
L37:
  mov    -0x10(%rbp),%rax
  mov    $0xa,%esi
  mov    %rax,%rdi
  call   <strchr@plt>
  mov    %rax,-0x8(%rbp)
  cmpq   $0x0,-0x8(%rbp)
  jne    L91
  mov    -0x10(%rbp),%rax
L99:
  leave
  ret
```

tool_calls: name: retrieve_data
arguments: tag: D0
datatype: string

tool: D0: < EMPTY >

assistant: ```c
char *abbreviate(const char *s, int len) {
  char *t, *p;
  int i, n, half;
  if (s == NULL)
    return my_strdup("< EMPTY >");
  n = (int)strlen(s);
  if (n < len) {
    t = my_strdup(s);
  } else {
    t = (char *)malloc(len * sizeof(char));
    half = (len - 6) / 2;
    for (i = 0; i <= half; i++)
      t[i] = s[i];
    t[half + 1] = ' ';
    t[half + 2] = '.';
    t[half + 3] = '.';
    t[half + 4] = '.';
    t[half + 5] = ' ';
    for (i = 0; i <= half; i++)
      t[half + 6 + i] = s[n - half + i];
  }
  while ((p = strchr(t, '\n')))
    *p = ' ';
  return t;
}
```