user: 下面的汇编指令对应的C源代码是什么：

```asm
<ConvertPath>:
  endbr64
  push   %rbp
  push   %rbx
  cmpb   $0x0,(%rsi)
  je     L202
  mov    %rsi,%r8
  xor    %eax,%eax
  nopl   0x0(%rax)
L115:
  mov    %eax,%edx
  add    $0x1,%rax
  cmpb   $0x0,(%r8,%rax,1)
  jne    L115
  lea    0x1(%rdx),%r10d
  xor    %ecx,%ecx
  xor    %edx,%edx
  mov    $0x1,%r9d
  mov    $0xba2e8ba3,%r11d
  jmp    L11
  nopl   0x0(%rax,%rax,1)
L149:
  cmp    $0x2e,%al
  je     L211
  movslq %edx,%rsi
  add    $0x1,%edx
  add    %rdi,%rsi
  cmp    $0xe5,%al
  je     L8
  lea    -0x61(%rax),%ebp
  lea    -0x20(%rax),%ebx
  cmp    $0x1a,%bpl
  cmovb  %ebx,%eax
  mov    %al,(%rsi)
  imul   $0xba2e8ba3,%edx,%eax
  cmp    $0x1745d174,%eax
  ja     L127
  movzbl 0x1(%r8,%rcx,1),%eax
  cmp    $0x2f,%al
  je     L127
  test   %al,%al
  jne    L52
  nopw   0x0(%rax,%rax,1)
L127:
  add    $0x1,%rcx
  cmp    %rcx,%r10
  je     L81
L11:
  movzbl (%r8,%rcx,1),%eax
  cmp    $0x2f,%al
  jne    L149
  test   %rcx,%rcx
  je     L127
  imul   $0xba2e8ba3,%edx,%eax
  cmp    $0x1745d174,%eax
  jbe    L51
  lea    0x1(%rdx),%eax
  cltq
  nopl   0x0(%rax)
L46:
  mov    %eax,%edx
  movb   $0x20,-0x1(%rdi,%rax,1)
  add    $0x1,%rax
  imul   $0xba2e8ba3,%edx,%esi
  cmp    $0x1745d174,%esi
  ja     L46
L51:
  add    $0x1,%rcx
  add    $0x1,%r9d
  cmp    %rcx,%r10
  jne    L11
  cs nopw 0x0(%rax,%rax,1)
L81:
  movslq %edx,%rcx
  add    %rdi,%rcx
L84:
  movzbl (%rcx),%eax
  mov    $0xffffffa3,%esi
  imul   %esi,%eax
  add    $0xb,%eax
  cmp    $0x16,%al
  jbe    L197
  movslq %edx,%rdx
  lea    0x1(%rdi,%rdx,1),%rdx
L217:
  movb   $0x20,(%rcx)
  mov    %rdx,%rcx
  add    $0x1,%rdx
  movzbl (%rcx),%eax
  imul   %esi,%eax
  add    $0xb,%eax
  cmp    $0x16,%al
  ja     L217
L197:
  movb   $0x0,(%rcx)
  mov    %r9d,%eax
  pop    %rbx
  pop    %rbp
  ret
L8:
  movb   $0x5,(%rsi)
  jmp    L127
  nopl   0x0(%rax,%rax,1)
L211:
  mov    %edx,%eax
  imul   %r11,%rax
  shr    $0x23,%rax
  lea    (%rax,%rax,4),%esi
  lea    (%rax,%rsi,2),%esi
  mov    %edx,%eax
  sub    %esi,%eax
  cmp    $0x8,%eax
  jg     L52
  test   $0x7,%al
  je     L127
  lea    0x1(%rdx),%esi
  movslq %esi,%rsi
  nopw   0x0(%rax,%rax,1)
L109:
  mov    %esi,%eax
  movb   $0x20,-0x1(%rdi,%rsi,1)
  add    $0x1,%rsi
  mov    %rax,%rdx
  imul   %r11,%rax
  shr    $0x23,%rax
  lea    (%rax,%rax,4),%ebx
  lea    (%rax,%rbx,2),%ebx
  mov    %edx,%eax
  sub    %ebx,%eax
  test   $0x7,%al
  jne    L109
  jmp    L127
L52:
  xor    %r9d,%r9d
  pop    %rbx
  pop    %rbp
  mov    %r9d,%eax
  ret
L202:
  mov    %rdi,%rcx
  xor    %edx,%edx
  mov    $0x1,%r9d
  jmp    L84
```

assistant: ```c
int ConvertPath(char *Out, char *In) {
  int depth = 1;
  int InLength = 0;
  int OutLength = 0;
  char *temp = In;
  while (*temp++)
    InLength++;
  for (int i = 0; i < InLength; i++) {
    if ((In[i] == '/')) {
      if (i) {
        while (OutLength % 11)
          Out[OutLength++] = ' ';
        depth++;
      }
    } else if (In[i] == '.') {
      if ((OutLength % 11) > 8)
        return 0;
      while ((OutLength % 11) % 8)
        Out[OutLength++] = ' ';
    } else if (In[i] == (char)0xE5) {
      Out[OutLength++] = 5;
    } else {
      if (In[i] >= 'a' && In[i] <= 'z')
        Out[OutLength++] = (char)(In[i] - ' ');
      else
        Out[OutLength++] = In[i];
      if (!(OutLength % 11) && ((In[i + 1] != '/') && (In[i + 1] != 0)))
        return 0;
    }
  }
  while ((Out[OutLength] % 11))
    Out[OutLength++] = ' ';
  Out[OutLength] = 0;
  return depth;
}
```