user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<reverseWords>:
  endbr64
  push   %rbp
  push   %rbx
  sub    $0x8,%rsp
  mov    %rdi,%rbx
  call   <strlen@plt>
  cmpb   $0x20,(%rbx)
  jne    L2
  mov    $0x1,%edx
L203:
  mov    %edx,%esi
  add    $0x1,%rdx
  cmpb   $0x20,-0x1(%rbx,%rdx,1)
  je     L203
L88:
  cmp    %esi,%eax
  jl     L243
  mov    %rbx,%rdx
  mov    %eax,%ecx
  sub    %esi,%ecx
  lea    0x1(%rbx,%rcx,1),%r8
  movslq %esi,%rdi
L153:
  movzbl (%rdx,%rdi,1),%ecx
  mov    %cl,(%rdx)
  add    $0x1,%rdx
  cmp    %r8,%rdx
  jne    L153
L243:
  sub    %esi,%eax
  movslq %eax,%rcx
  cmpb   $0x20,-0x1(%rbx,%rcx,1)
  jne    L231
  sub    $0x1,%eax
  movslq %eax,%rdx
L223:
  mov    %edx,%eax
  mov    %rdx,%rcx
  sub    $0x1,%rdx
  cmpb   $0x20,(%rbx,%rdx,1)
  je     L223
L231:
  movb   $0x0,(%rbx,%rcx,1)
  test   %eax,%eax
  jle    L102
  mov    %rbx,%rdx
  lea    -0x1(%rax),%eax
  lea    0x1(%rbx,%rax,1),%rsi
  mov    $0x0,%r9d
  mov    $0x0,%ecx
  jmp    L186
L2:
  mov    $0x0,%esi
  jmp    L88
L91:
  test   %ecx,%ecx
  jne    L62
  movslq %r9d,%rax
  movb   $0x20,(%rbx,%rax,1)
  lea    0x1(%r9),%r9d
L62:
  add    $0x1,%ecx
L233:
  add    $0x1,%rdx
  cmp    %rsi,%rdx
  je     L32
L186:
  movzbl (%rdx),%eax
  cmp    $0x20,%al
  je     L91
  movslq %r9d,%rcx
  mov    %al,(%rbx,%rcx,1)
  lea    0x1(%r9),%r9d
  mov    $0x0,%ecx
  jmp    L233
L32:
  movslq %r9d,%rax
  movb   $0x0,(%rbx,%rax,1)
  mov    %r9d,%edi
  shr    $0x1f,%edi
  add    %r9d,%edi
  sar    %edi
  cmp    $0x1,%r9d
  jle    L31
  lea    -0x1(%rbx,%rax,1),%rdx
  mov    $0x0,%eax
L7:
  movzbl (%rbx,%rax,1),%ecx
  movzbl (%rdx),%esi
  mov    %sil,(%rbx,%rax,1)
  mov    %cl,(%rdx)
  add    $0x1,%rax
  sub    $0x1,%rdx
  cmp    %eax,%edi
  jg     L7
L117:
  mov    %r9d,%ebp
  mov    $0x0,%ecx
  mov    $0x0,%esi
  sub    $0x1,%r9d
  jmp    L188
L31:
  test   %r9d,%r9d
  jg     L117
  jmp    L183
L176:
  cmp    %ecx,%r9d
  je     L139
  lea    -0x1(%rcx),%r11d
L139:
  mov    %r11d,%eax
  sub    %esi,%eax
  lea    0x1(%rax),%edx
  mov    %edx,%r10d
  shr    $0x1f,%r10d
  add    %edx,%r10d
  sar    %r10d
  test   %eax,%eax
  jle    L57
  movslq %r11d,%rdx
  add    %rbx,%rdx
  mov    $0x0,%eax
  movslq %esi,%rsi
  add    %rbx,%rsi
L110:
  movzbl (%rsi,%rax,1),%edi
  movzbl (%rdx),%r8d
  mov    %r8b,(%rsi,%rax,1)
  mov    %dil,(%rdx)
  add    $0x1,%rax
  sub    $0x1,%rdx
  cmp    %eax,%r10d
  jg     L110
L57:
  lea    0x2(%r11),%esi
L44:
  add    $0x1,%rcx
  cmp    %rbp,%rcx
  je     L183
L188:
  mov    %ecx,%r11d
  cmpb   $0x20,(%rbx,%rcx,1)
  je     L176
  cmp    %ecx,%r9d
  jne    L44
  jmp    L139
L102:
  movb   $0x0,(%rbx)
L183:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
```

assistant: ```c
void reverseWords(char *s) {
  int len = strlen(s);
  int count = 0;
  while (s[count] == ' ')
    count++;
  for (int i = count; i < len + 1; i++)
    s[i - count] = s[i];
  len -= count;
  while (s[len - 1] == ' ')
    len--;
  s[len] = '\0';
  int index = 0;
  count = 0;
  for (int i = 0; i < len; i++) {
    if (s[i] != ' ') {
      s[index++] = s[i];
      count = 0;
    } else {
      if (!count)
        s[index++] = s[i];
      count++;
    }
  }
  s[index] = '\0';
  len = index;
  for (int i = 0; i < len / 2; i++) {
    char c = s[i];
    s[i] = s[len - i - 1];
    s[len - i - 1] = c;
  }
  int begin = 0, end = 0;
  for (int i = 0; i < len; i++) {
    if (s[i] == ' ' || i == len - 1) {
      end = i;
      if (i != len - 1)
        end--;
      int size = end - begin + 1;
      for (int j = 0; j < size / 2; j++) {
        char c = s[begin + j];
        s[begin + j] = s[end - j];
        s[end - j] = c;
      }
      begin = end + 2;
    }
  }
}
```