user: 这段汇编代码用C语言表示是什么样的？

```asm
<mystrtok>:
  endbr64
  push   %r13
  mov    %rsi,%r13
  push   %r12
  push   %rbp
  mov    %rdi,%rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    (%rdi),%r12
  movsbl (%r12),%esi
  test   %sil,%sil
  jne    L53
  jmp    L3
  nopl   0x0(%rax,%rax,1)
L59:
  add    $0x1,%r12
  mov    %r12,0x0(%rbp)
  movsbl (%r12),%esi
  test   %sil,%sil
  je     L3
L53:
  mov    %r13,%rdi
  call   <strchr@plt>
  test   %rax,%rax
  jne    L59
  mov    0x0(%rbp),%rbx
  nopl   0x0(%rax,%rax,1)
L45:
  movsbl (%rbx),%esi
  test   %sil,%sil
  je     L14
  mov    %r13,%rdi
  call   <strchr@plt>
  mov    %rbx,%rdx
  add    $0x1,%rbx
  mov    %rbx,0x0(%rbp)
  test   %rax,%rax
  je     L45
  movb   $0x0,(%rdx)
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
  nopl   0x0(%rax)
L3:
  xor    %r12d,%r12d
L14:
  add    $0x8,%rsp
  mov    %r12,%rax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  ret
```

assistant: ```c
char *mystrtok(char **ptr, char const *sep) {
  char *res;
  char *tmp;
  char *tmp___0;
  char *tmp___1;
  char *__cil_tmp7;
  char __cil_tmp8;
  int __cil_tmp9;
  void *__cil_tmp10;
  char *__cil_tmp11;
  char *__cil_tmp12;
  char __cil_tmp13;
  int __cil_tmp14;
  char *__cil_tmp15;
  char *__cil_tmp16;
  char __cil_tmp17;
  int __cil_tmp18;
  void *__cil_tmp19;
  char *__cil_tmp20;
  char *__cil_tmp21;
  char __cil_tmp22;
  int __cil_tmp23;
  void *__cil_tmp24;
  unsigned long __cil_tmp25;
  unsigned long __cil_tmp26;
  char *__cil_tmp27;
  char *__cil_tmp28;
  char __cil_tmp29;
  int __cil_tmp30;
  char *__cil_tmp31;
  {
    {
      __cil_tmp7 = *ptr;
      __cil_tmp8 = *__cil_tmp7;
      __cil_tmp9 = (int)__cil_tmp8;
      if (__cil_tmp9 == 0) {
        {
          __cil_tmp10 = (void *)0;
          return ((char *)__cil_tmp10);
        }
      } else {
      }
    }
    {
      while (1) {
      while_13_continue:;
        {
          __cil_tmp11 = *ptr;
          if (*__cil_tmp11) {
            {
              __cil_tmp12 = *ptr;
              __cil_tmp13 = *__cil_tmp12;
              __cil_tmp14 = (int)__cil_tmp13;
              tmp = strchr(sep, __cil_tmp14);
            }
            if (tmp) {
            } else {
              goto while_13_break;
            }
          } else {
            goto while_13_break;
          }
        }
        __cil_tmp15 = *ptr;
        *ptr = __cil_tmp15 + 1;
      }
    while_13_break:;
    }
    {
      __cil_tmp16 = *ptr;
      __cil_tmp17 = *__cil_tmp16;
      __cil_tmp18 = (int)__cil_tmp17;
      if (__cil_tmp18 == 0) {
        {
          __cil_tmp19 = (void *)0;
          return ((char *)__cil_tmp19);
        }
      } else {
      }
    }
    res = *ptr;
    {
      while (1) {
      while_14_continue:;
        {
          __cil_tmp20 = *ptr;
          if (*__cil_tmp20) {
            {
              __cil_tmp21 = *ptr;
              __cil_tmp22 = *__cil_tmp21;
              __cil_tmp23 = (int)__cil_tmp22;
              tmp___0 = strchr(sep, __cil_tmp23);
            }
            {
              __cil_tmp24 = (void *)0;
              __cil_tmp25 = (unsigned long)__cil_tmp24;
              __cil_tmp26 = (unsigned long)tmp___0;
              if (__cil_tmp26 == __cil_tmp25) {
              } else {
                goto while_14_break;
              }
            }
          } else {
            goto while_14_break;
          }
        }
        __cil_tmp27 = *ptr;
        *ptr = __cil_tmp27 + 1;
      }
    while_14_break:;
    }
    {
      __cil_tmp28 = *ptr;
      __cil_tmp29 = *__cil_tmp28;
      __cil_tmp30 = (int)__cil_tmp29;
      if (__cil_tmp30 != 0) {
        tmp___1 = *ptr;
        __cil_tmp31 = *ptr;
        *ptr = __cil_tmp31 + 1;
        *tmp___1 = (char)0;
      } else {
      }
    }
    return (res);
  }
}
```