user: 这段汇编代码用C语言表示是什么样的?

```asm
<my_lltoa>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x30,%rsp
  mov    %rdi,-0x18(%rbp)
  mov    %rsi,-0x20(%rbp)
  mov    %edx,-0x24(%rbp)
  mov    %ecx,-0x28(%rbp)
  movl   $0x0,-0x8(%rbp)
  subl   $0x1,-0x24(%rbp)
  mov    -0x28(%rbp),%eax
  cmp    -0x24(%rbp),%eax
  jle    L63
  mov    -0x24(%rbp),%edx
  mov    -0x28(%rbp),%eax
  mov    %eax,%esi
  lea    D4(%rip),%rax
  mov    %rax,%rdi
  call   <debuga@plt>
  mov    $0x0,%eax
  call   <abort@plt>
L63:
  mov    -0x18(%rbp),%rcx
  movabs $0xcccccccccccccccd,%rdx
  mov    %rcx,%rax
  mul    %rdx
  shr    $0x3,%rdx
  mov    %rdx,%rax
  shl    $0x2,%rax
  add    %rdx,%rax
  add    %rax,%rax
  sub    %rax,%rcx
  mov    %rcx,%rdx
  mov    %edx,%eax
  lea    0x30(%rax),%ecx
  mov    -0x8(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x8(%rbp)
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    %ecx,%edx
  mov    %dl,(%rax)
  mov    -0x18(%rbp),%rax
  movabs $0xcccccccccccccccd,%rdx
  mul    %rdx
  mov    %rdx,%rax
  shr    $0x3,%rax
  mov    %rax,-0x18(%rbp)
  cmpq   $0x0,-0x18(%rbp)
  je     L28
  mov    -0x8(%rbp),%eax
  cmp    -0x24(%rbp),%eax
  jl     L63
L28:
  mov    -0x8(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  movl   $0x0,-0xc(%rbp)
  mov    -0x8(%rbp),%eax
  sub    $0x1,%eax
  mov    %eax,-0x4(%rbp)
  jmp    L90
L76:
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  mov    %al,-0xd(%rbp)
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    -0xc(%rbp),%edx
  movslq %edx,%rcx
  mov    -0x20(%rbp),%rdx
  add    %rcx,%rdx
  movzbl (%rax),%eax
  mov    %al,(%rdx)
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rax,%rdx
  movzbl -0xd(%rbp),%eax
  mov    %al,(%rdx)
  addl   $0x1,-0xc(%rbp)
  subl   $0x1,-0x4(%rbp)
L90:
  mov    -0xc(%rbp),%eax
  cmp    -0x4(%rbp),%eax
  jl     L76
  mov    -0x28(%rbp),%eax
  cmp    -0x8(%rbp),%eax
  jle    L47
  mov    -0x28(%rbp),%eax
  sub    -0x8(%rbp),%eax
  mov    %eax,-0xc(%rbp)
  mov    -0x8(%rbp),%eax
  mov    %eax,-0x4(%rbp)
  jmp    L92
L11:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    -0x4(%rbp),%ecx
  mov    -0xc(%rbp),%edx
  add    %ecx,%edx
  movslq %edx,%rcx
  mov    -0x20(%rbp),%rdx
  add    %rcx,%rdx
  movzbl (%rax),%eax
  mov    %al,(%rdx)
  subl   $0x1,-0x4(%rbp)
L92:
  cmpl   $0x0,-0x4(%rbp)
  jns    L11
  movl   $0x0,-0x4(%rbp)
  jmp    L78
L79:
  mov    -0x4(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movb   $0x30,(%rax)
  addl   $0x1,-0x4(%rbp)
L78:
  mov    -0x4(%rbp),%eax
  cmp    -0xc(%rbp),%eax
  jl     L79
L47:
  nop
  leave
  ret
```

tool_calls: name: read_data
arguments: data_tag: D4
datatype: string

tool: D4: The requested number of digits passed to my_lltoa (%d) is bigger than the output buffer size (%d)


assistant: ```c
void my_lltoa(unsigned long long int n, char *s, int ssize, int len) {
  int i;
  int slen = 0;
  int j;
  char c;
  ssize--;
  if (len > ssize) {
    debuga(("The requested number of digits passed to my_lltoa (%d) is bigger than the output buffer size (%d)\n"), len, ssize);
    abort();
  }
  do {
    s[slen++] = (n % 10) + '0';
  } while ((n /= 10) > 0 && slen < ssize);
  s[slen] = '\0';
  for (i = 0, j = slen - 1; i < j; i++, j--) {
    c = s[i];
    s[i] = s[j];
    s[j] = c;
  }
  if (len > slen) {
    i = len - slen;
    for (j = slen; j >= 0; j--)
      s[j + i] = s[j];
    for (j = 0; j < i; j++)
      s[j] = '0';
  }
}
```