user: 请从给定的汇编指令推导出C源代码：

```asm
<my_str_to_wordtab>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  push   %rbx
  sub    $0x38,%rsp
  mov    %rdi,-0x38(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x18(%rbp)
  xor    %eax,%eax
  movl   $0x0,-0x2c(%rbp)
  movl   $0x0,-0x28(%rbp)
  mov    -0x38(%rbp),%rax
  mov    %rax,%rdi
  call   <my_strdup@plt>
  mov    %rax,%rdi
  call   <my_count_word@plt>
  mov    %eax,-0x24(%rbp)
  mov    -0x24(%rbp),%eax
  cltq
  shl    $0x3,%eax
  mov    %eax,%edi
  call   <malloc@plt>
  mov    %rax,-0x20(%rbp)
  jmp    L28
L72:
  mov    -0x2c(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x38(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x60,%al
  jle    L97
  mov    -0x2c(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x38(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x7a,%al
  jle    L68
L97:
  mov    -0x2c(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x38(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x40,%al
  jle    L58
  mov    -0x2c(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x38(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x5a,%al
  jle    L68
L58:
  mov    -0x2c(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x38(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x2f,%al
  jle    L3
  mov    -0x2c(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x38(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x39,%al
  jg     L3
L68:
  mov    -0x2c(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x38(%rbp),%rax
  add    %rdx,%rax
  mov    -0x28(%rbp),%edx
  movslq %edx,%rdx
  lea    0x0(,%rdx,8),%rcx
  mov    -0x20(%rbp),%rdx
  lea    (%rcx,%rdx,1),%rbx
  mov    %rax,%rdi
  call   <my_strdup@plt>
  mov    %rax,(%rbx)
  mov    -0x28(%rbp),%eax
  cltq
  lea    0x0(,%rax,8),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  mov    (%rax),%rbx
  lea    -0x2c(%rbp),%rdx
  mov    -0x38(%rbp),%rax
  mov    %rdx,%rsi
  mov    %rax,%rdi
  call   <my_count_char@plt>
  add    %rbx,%rax
  movb   $0x0,(%rax)
  addl   $0x1,-0x28(%rbp)
  subl   $0x1,-0x24(%rbp)
L3:
  mov    -0x2c(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,-0x2c(%rbp)
L28:
  mov    -0x2c(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x38(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  je     L20
  cmpl   $0x0,-0x24(%rbp)
  jg     L72
L20:
  mov    -0x38(%rbp),%rax
  mov    %rax,%rdi
  call   <my_strdup@plt>
  mov    %rax,%rdi
  call   <my_count_word@plt>
  lea    0x0(,%rax,8),%rdx
  mov    -0x20(%rbp),%rax
  add    %rdx,%rax
  movq   $0x0,(%rax)
  mov    -0x20(%rbp),%rax
  mov    -0x18(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L16
  call   <__stack_chk_fail@plt>
L16:
  mov    -0x8(%rbp),%rbx
  leave
  ret
```

assistant: ```c
char **my_str_to_wordtab(char *str) {
  int i;
  int n;
  int word;
  char **tab;
  i = 0;
  n = 0;
  word = my_count_word(my_strdup(str));
  tab = (char **)malloc(sizeof(char *) * word);
  while (str[i] && word > 0) {
    if ((str[i] >= 'a' && str[i] <= 'z') || (str[i] >= 'A' && str[i] <= 'Z') || (str[i] >= '0' && str[i] <= '9')) {
      tab[n] = my_strdup(str + i);
      tab[n][my_count_char(str, &i)] = '\0';
      n++;
      word--;
    }
    i++;
  }
  tab[my_count_word(my_strdup(str))] = NULL;
  return (tab);
}
```