user: 请从给定的汇编指令推导出C源代码:

```asm
<od>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %r15
  mov    %rsi,%r15
  push   %r14
  mov    %rdi,%r14
  mov    %ecx,%edi
  push   %r13
  mov    %ecx,%r13d
  push   %r12
  mov    %rdx,%r12
  push   %rbx
  sub    $0x58,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x38(%rbp)
  xor    %eax,%eax
  call   <_length@plt>
  mov    %rax,%rdi
  mov    %rax,%rbx
  call   <_log10@plt>
  mov    %rsp,%rcx
  add    $0x7,%eax
  cltq
  add    $0xf,%rax
  mov    %rax,%rdx
  and    $0xfffffffffffff000,%rax
  sub    %rax,%rcx
  and    $0xfffffffffffffff0,%rdx
  cmp    %rcx,%rsp
  je     L78
L118:
  sub    $0x1000,%rsp
  orq    $0x0,0xff8(%rsp)
  cmp    %rcx,%rsp
  jne    L118
L78:
  and    $0xfff,%edx
  sub    %rdx,%rsp
  test   %rdx,%rdx
  jne    L113
L125:
  lea    -0x3d(%rbp),%rax
  mov    %r13d,%esi
  mov    %rsp,-0x60(%rbp)
  mov    %rax,%rdi
  mov    %rax,-0x78(%rbp)
  call   <_address_format@plt>
  mov    %r13d,%esi
  mov    %rsp,%rdi
  call   <_bytes_format@plt>
  test   %r12,%r12
  mov    $0x10,%eax
  movq   $0x1,-0x50(%rbp)
  cmove  %rax,%r12
  cmp    %rbx,%r12
  jae    L43
L39:
  xor    %edx,%edx
  mov    %r15,%rax
  div    %rbx
  cmp    $0x1,%rdx
  sbb    $0xffffffffffffffff,%rax
  mov    %rax,-0x48(%rbp)
  mov    %r13d,%eax
  and    $0xc000,%eax
  mov    %eax,-0x6c(%rbp)
  cmp    $0xc000,%eax
  je     L139
  mov    -0x78(%rbp),%rdi
  xor    %esi,%esi
  xor    %eax,%eax
  call   <printf@plt>
L139:
  mov    -0x48(%rbp),%rax
  mov    %rbx,%r12
  xor    %r15d,%r15d
  lea    -0x1(%rax),%rdi
  mov    %rdi,-0x58(%rbp)
  test   %rax,%rax
  je     L37
  mov    %r13d,%eax
  mov    %r15,%r13
  mov    %eax,%r15d
  jmp    L88
  nopw   0x0(%rax,%rax,1)
L89:
  cmp    %rcx,-0x58(%rbp)
  jne    L40
  mov    $0xa,%edi
  call   <putchar@plt>
L40:
  add    %rbx,%r12
  cmp    -0x48(%rbp),%r13
  je     L37
L88:
  mov    -0x60(%rbp),%rdx
  mov    %rbx,%rcx
  mov    %r13,%rsi
  mov    %r15d,%r8d
  mov    %r14,%rdi
  call   <_print_date@plt>
  mov    %r13,%rcx
  add    $0x1,%r13
  xor    %edx,%edx
  mov    %r13,%rax
  divq   -0x50(%rbp)
  test   %rdx,%rdx
  jne    L89
  mov    $0xa,%edi
  mov    %rcx,-0x68(%rbp)
  call   <putchar@plt>
  cmpl   $0xc000,-0x6c(%rbp)
  je     L40
  mov    -0x68(%rbp),%rcx
  cmp    %rcx,-0x58(%rbp)
  je     L40
  mov    -0x78(%rbp),%rdi
  mov    %r12,%rsi
  xor    %eax,%eax
  add    %rbx,%r12
  call   <printf@plt>
  cmp    -0x48(%rbp),%r13
  jne    L88
  nopl   0x0(%rax,%rax,1)
L37:
  mov    -0x38(%rbp),%rax
  sub    %fs:0x28,%rax
  jne    L65
  lea    -0x28(%rbp),%rsp
  pop    %rbx
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  pop    %rbp
  ret
  cs nopw 0x0(%rax,%rax,1)
L43:
  mov    %r12,%rax
  xor    %edx,%edx
  div    %rbx
  mov    %rax,-0x50(%rbp)
  jmp    L39
  nopl   0x0(%rax)
L113:
  orq    $0x0,-0x8(%rsp,%rdx,1)
  jmp    L125
L65:
  call   <__stack_chk_fail@plt>
```

assistant: ```c
void od(void *data, size_t data_len, uint8_t width, uint16_t flags) {
  char address_format[5];
  uint8_t date_length = _length(flags);
  char bytes_format[_log10(date_length) + 7];
  _address_format(address_format, flags);
  _bytes_format(bytes_format, flags);
  if (width == 0) {
    width = (16);
  }
  if (width < date_length) {
    width = 1;
  } else {
    width = (width / date_length);
  }
  if (data_len % date_length) {
    data_len = (data_len / date_length) + 1;
  } else {
    data_len = data_len / date_length;
  }
  if ((flags & (0xc000)) != (0xc000)) {
    printf(address_format, 0);
  }
  for (size_t i = 0; i < data_len; i++) {
    _print_date(data, i, bytes_format, date_length, flags);
    if ((((i + 1) % width) == 0) || i == (data_len - 1)) {
      printf("\n");
      if (i != (data_len - 1)) {
        if ((flags & (0xc000)) != (0xc000)) {
          printf(address_format, date_length * (i + 1));
        }
      }
    }
  }
}
```