user: What would the C source look like for the assembly code provided here:

```asm
<reverseWords>:
  endbr64
  push   %rbp
  push   %rbx
  mov    %rdi,%rbx
  sub    $0x8,%rsp
  call   <strlen@plt>
  cmpb   $0x20,(%rbx)
  jne    L192
  mov    $0x1,%edx
L10:
  movslq %edx,%rcx
  add    $0x1,%rdx
  cmpb   $0x20,-0x1(%rbx,%rdx,1)
  je     L10
  mov    %eax,%esi
  sub    %ecx,%esi
L95:
  cmp    %ecx,%eax
  jl     L161
  sub    %ecx,%eax
  mov    %rbx,%rdx
  lea    0x1(%rbx,%rax,1),%rdi
L208:
  movzbl (%rdx,%rcx,1),%eax
  add    $0x1,%rdx
  mov    %al,-0x1(%rdx)
  cmp    %rdi,%rdx
  jne    L208
L161:
  movslq %esi,%rdx
  cmpb   $0x20,-0x1(%rbx,%rdx,1)
  lea    -0x1(%rdx),%rax
  jne    L108
  xchg   %ax,%ax
L189:
  mov    %eax,%esi
  mov    %rax,%rdx
  sub    $0x1,%rax
  cmpb   $0x20,(%rbx,%rax,1)
  je     L189
L108:
  movb   $0x0,(%rbx,%rdx,1)
  test   %esi,%esi
  jle    L142
  lea    -0x1(%rsi),%eax
  mov    %rbx,%rdx
  xor    %edi,%edi
  xor    %r8d,%r8d
  lea    0x1(%rbx,%rax,1),%r9
  xor    %r10d,%r10d
  mov    %rbx,%rax
  jmp    L234
  nopl   0x0(%rax,%rax,1)
L244:
  add    $0x1,%edi
  add    $0x1,%rax
  mov    %cl,(%rsi)
  xor    %r8d,%r8d
  movslq %edi,%r10
  cmp    %rax,%r9
  je     L230
L234:
  movzbl (%rax),%ecx
  lea    (%rbx,%r10,1),%rsi
  cmp    $0x20,%cl
  jne    L244
  test   %r8d,%r8d
  jne    L78
  movb   $0x20,(%rsi)
  add    $0x1,%edi
  movslq %edi,%r10
L78:
  add    $0x1,%rax
  add    $0x1,%r8d
  cmp    %rax,%r9
  jne    L234
L230:
  mov    %edi,%ecx
  movb   $0x0,(%rbx,%r10,1)
  sar    %ecx
  je     L68
  lea    -0x2(%rbx,%r10,1),%r8
  sub    $0x1,%ecx
  lea    -0x1(%rbx,%r10,1),%rax
  sub    %rcx,%r8
  nopw   0x0(%rax,%rax,1)
L110:
  movzbl (%rdx),%ecx
  movzbl (%rax),%esi
  sub    $0x1,%rax
  add    $0x1,%rdx
  mov    %sil,-0x1(%rdx)
  mov    %cl,0x1(%rax)
  cmp    %rax,%r8
  jne    L110
L138:
  xor    %esi,%esi
  xor    %ecx,%ecx
  lea    -0x1(%rdi),%ebp
  jmp    L169
  nopl   0x0(%rax,%rax,1)
L82:
  cmp    %esi,%ebp
  je     L34
  add    $0x1,%rsi
  cmp    %rsi,%r10
  je     L112
L169:
  cmpb   $0x20,(%rbx,%rsi,1)
  mov    %esi,%r11d
  jne    L82
  cmp    %esi,%ebp
  je     L34
  sub    $0x1,%r11d
L34:
  mov    %r11d,%eax
  sub    %ecx,%eax
  lea    0x1(%rax),%edx
  mov    %edx,%r9d
  shr    $0x1f,%r9d
  add    %edx,%r9d
  sar    %r9d
  test   %eax,%eax
  jle    L94
  movslq %r11d,%rdx
  movslq %ecx,%rcx
  xor    %eax,%eax
  add    %rbx,%rdx
  add    %rbx,%rcx
  nopl   (%rax)
L66:
  movzbl (%rdx),%r8d
  movzbl (%rcx,%rax,1),%edi
  sub    $0x1,%rdx
  mov    %r8b,(%rcx,%rax,1)
  add    $0x1,%rax
  mov    %dil,0x1(%rdx)
  cmp    %eax,%r9d
  jg     L66
L94:
  add    $0x1,%rsi
  lea    0x2(%r11),%ecx
  cmp    %rsi,%r10
  jne    L169
L112:
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L192:
  mov    %eax,%esi
  xor    %ecx,%ecx
  jmp    L95
L142:
  movb   $0x0,(%rbx)
  add    $0x8,%rsp
  pop    %rbx
  pop    %rbp
  ret
L68:
  cmp    $0x1,%edi
  jne    L112
  jmp    L138
```

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;
    }
  }
}
```