user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<reverseWords>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x20,%rsp
  mov    %rdi,-0x18(%rbp)
  movl   $0x0,-0x10(%rbp)
  movl   $0x0,-0xc(%rbp)
  jmp    L15
L19:
  addl   $0x1,-0xc(%rbp)
L15:
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x20,%al
  je     L19
  jmp    L74
L131:
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x20,%al
  je     L111
  mov    -0xc(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0xc(%rbp)
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  lea    (%rdx,%rax,1),%rcx
  mov    -0x10(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x10(%rbp)
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  movzbl (%rcx),%eax
  mov    %al,(%rdx)
  jmp    L74
L111:
  mov    -0xc(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0xc(%rbp)
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  lea    (%rdx,%rax,1),%rcx
  mov    -0x10(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x10(%rbp)
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rax,%rdx
  movzbl (%rcx),%eax
  mov    %al,(%rdx)
  jmp    L37
L86:
  addl   $0x1,-0xc(%rbp)
L37:
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x20,%al
  je     L86
L74:
  mov    -0xc(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  test   %al,%al
  jne    L131
  mov    -0x10(%rbp),%eax
  cltq
  lea    -0x1(%rax),%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x20,%al
  jne    L99
  subl   $0x1,-0x10(%rbp)
L99:
  mov    -0x10(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  mov    -0x18(%rbp),%rax
  mov    %rax,%rdi
  call   <strlen@plt>
  mov    %eax,-0x4(%rbp)
  mov    -0x4(%rbp),%edx
  mov    -0x18(%rbp),%rax
  mov    %edx,%esi
  mov    %rax,%rdi
  call   <reverse@plt>
  movl   $0x0,-0xc(%rbp)
  jmp    L27
L58:
  mov    -0xc(%rbp),%eax
  mov    %eax,-0x8(%rbp)
  jmp    L30
L41:
  addl   $0x1,-0x8(%rbp)
L30:
  mov    -0x8(%rbp),%eax
  cmp    -0x4(%rbp),%eax
  jge    L56
  mov    -0x8(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x18(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  cmp    $0x20,%al
  jne    L41
L56:
  mov    -0x8(%rbp),%eax
  sub    -0xc(%rbp),%eax
  mov    -0xc(%rbp),%edx
  movslq %edx,%rcx
  mov    -0x18(%rbp),%rdx
  add    %rcx,%rdx
  mov    %eax,%esi
  mov    %rdx,%rdi
  call   <reverse@plt>
  mov    -0x8(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,-0xc(%rbp)
L27:
  mov    -0xc(%rbp),%eax
  cmp    -0x4(%rbp),%eax
  jl     L58
  nop
  nop
  leave
  ret
```

assistant: ```c
void reverseWords(char *s) {
  int k = 0;
  int i = 0;
  while (s[i] == ' ')
    ++i;
  while (s[i] != '\0')
    if (s[i] != ' ')
      s[k++] = s[i++];
    else {
      s[k++] = s[i++];
      while (s[i] == ' ')
        ++i;
    }
  if (s[k - 1] == ' ')
    --k;
  s[k] = '\0';
  int s_len = strlen(s);
  reverse(s, s_len);
  for (i = 0; i < s_len;) {
    int j = i;
    while (j < s_len && s[j] != ' ')
      ++j;
    reverse(s + i, j - i);
    i = j + 1;
  }
}
```