user: 请将这段汇编代码片段转换成等效的C代码：

```asm
<readlines>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x428,%rsp
  mov    %rdi,-0x428(%rbp)
  mov    %esi,-0x42c(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x18(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x418(%rbp)
  movl   $0x0,-0x410(%rbp)
  movl   $0x0,-0x40c(%rbp)
  movq   $0x0,-0x408(%rbp)
  jmp    L65
L63:
  addq   $0x1,-0x408(%rbp)
  mov    -0x418(%rbp),%eax
  add    $0x1,%eax
  cltd
  idivl  -0x42c(%rbp)
  mov    %edx,-0x418(%rbp)
  mov    -0x418(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x428(%rbp),%rax
  lea    (%rdx,%rax,1),%rbx
  mov    -0x40c(%rbp),%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,(%rbx)
  mov    -0x418(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x428(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  lea    -0x400(%rbp),%rdx
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <strcpy@plt>
L65:
  lea    -0x400(%rbp),%rax
  mov    $0x3e8,%esi
  mov    %rax,%rdi
  call   <mygetline@plt>
  mov    %eax,-0x40c(%rbp)
  cmpl   $0x0,-0x40c(%rbp)
  jg     L63
  mov    -0x42c(%rbp),%edx
  mov    -0x408(%rbp),%rax
  mov    %rax,%rsi
  lea    D3(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  mov    -0x42c(%rbp),%eax
  cltq
  cmp    %rax,-0x408(%rbp)
  jg     L36
  movl   $0x0,-0x418(%rbp)
  jmp    L8
L11:
  mov    -0x418(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x428(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  mov    %rax,%rsi
  lea    D25(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x418(%rbp)
L8:
  mov    -0x418(%rbp),%eax
  cltq
  cmp    %rax,-0x408(%rbp)
  jg     L11
  jmp    L2
L36:
  mov    -0x42c(%rbp),%eax
  mov    %eax,-0x414(%rbp)
  jmp    L23
L40:
  mov    -0x418(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x428(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rdx
  mov    -0x418(%rbp),%eax
  mov    %eax,%esi
  lea    D27(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  addl   $0x1,-0x418(%rbp)
  mov    -0x418(%rbp),%eax
  cltd
  idivl  -0x42c(%rbp)
  mov    %edx,%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x428(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rax
  mov    %rax,%rsi
  lea    D25(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L23:
  mov    -0x414(%rbp),%eax
  lea    -0x1(%rax),%edx
  mov    %edx,-0x414(%rbp)
  test   %eax,%eax
  jg     L40
L2:
  mov    -0x408(%rbp),%rax
  mov    -0x18(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L51
  call   <__stack_chk_fail@plt>
L51:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

tool_calls: name: unpack_data
arguments: data_label: D3
type_infered: string
name: unpack_data
arguments: data_label: D25
type_infered: string
name: unpack_data
arguments: data_label: D27
type_infered: string

tool: D3: Found lines %d || %d


tool: D25: %s

tool: D27: ***i is[%d]: (((%s)))


assistant: ```c
int readlines(char *lines[], int numlines) {
  int i = 0;
  int min = 0;
  int j;
  int len = 0;
  char line[1000];
  long lineno = 0;
  while ((len = mygetline(line, 1000)) > 0) {
    lineno++;
    i = (i + 1) % numlines;
    lines[i] = malloc(len);
    strcpy(lines[i], line);
  }
  printf("Found lines %d || %d\n", lineno, numlines);
  if (lineno <= numlines) {
    for (i = 0; i < lineno; i++) {
      printf("%s", lines[i]);
    }
  } else {
    j = numlines;
    while (j-- > 0) {
      printf("***i is[%d]: (((%s)))\n", i, lines[i]);
      printf("%s", lines[++i % numlines]);
    }
  }
  return lineno;
}
```