user: 这段汇编代码用C语言表示是什么样的?

```asm
<t1_init_ctxno_zc>:
  endbr64
  mov    %edi,%eax
  mov    %esi,%edi
  mov    %eax,%ecx
  mov    %eax,%edx
  mov    %eax,%esi
  mov    %eax,%r8d
  shr    $0x7,%ecx
  shr    $0x5,%edx
  and    $0x1,%r8d
  and    $0x1,%edx
  and    $0x1,%ecx
  shr    $0x4,%esi
  add    %edx,%ecx
  mov    %eax,%edx
  and    $0x1,%esi
  shr    $0x6,%edx
  and    $0x1,%edx
  add    %edx,%esi
  mov    %eax,%edx
  shr    $0x3,%edx
  and    $0x1,%edx
  add    %r8d,%edx
  mov    %eax,%r8d
  shr    $0x2,%eax
  shr    %r8d
  and    $0x1,%eax
  and    $0x1,%r8d
  add    %r8d,%edx
  add    %eax,%edx
  cmp    $0x2,%edi
  je     L95
  jg     L5
  mov    $0x1,%eax
  cmp    $0x1,%edi
  ja     L29
L69:
  test   %ecx,%ecx
  jne    L84
  test   %esi,%esi
  jne    L27
  mov    $0x1,%eax
  test   %edx,%edx
  je     L62
  xor    %eax,%eax
  cmp    $0x1,%edx
  setne  %al
  add    $0x2,%eax
  ret
  nopl   (%rax)
L5:
  cmp    $0x3,%edi
  jne    L64
  add    %esi,%ecx
  test   %edx,%edx
  jne    L88
  mov    $0x1,%eax
  test   %ecx,%ecx
  je     L62
  xor    %eax,%eax
  cmp    $0x1,%ecx
  setne  %al
  add    $0x2,%eax
  ret
  nopl   0x0(%rax,%rax,1)
L84:
  mov    $0x9,%eax
  cmp    $0x1,%ecx
  je     L54
L62:
  ret
  nopl   0x0(%rax,%rax,1)
L64:
  mov    $0x1,%eax
  ret
  cs nopw 0x0(%rax,%rax,1)
L29:
  ret
  nopl   0x0(%rax)
L95:
  mov    %ecx,%eax
  mov    %esi,%ecx
  mov    %eax,%esi
  jmp    L69
L27:
  xor    %eax,%eax
  cmp    $0x1,%esi
  setne  %al
  add    $0x4,%eax
  ret
  nopl   0x0(%rax)
L88:
  cmp    $0x1,%edx
  je     L119
  cmp    $0x1,%ecx
  mov    $0x7,%eax
  sbb    $0xffffffff,%eax
  cmp    $0x2,%edx
  mov    $0x9,%edx
  cmovne %edx,%eax
  ret
  nopl   0x0(%rax)
L54:
  cmp    $0x1,%edx
  mov    $0x6,%eax
  mov    $0x8,%edx
  sbb    $0xffffffff,%eax
  test   %esi,%esi
  cmovne %edx,%eax
  ret
  cs nopw 0x0(%rax,%rax,1)
L119:
  mov    $0x4,%eax
  test   %ecx,%ecx
  je     L62
  xor    %eax,%eax
  cmp    $0x1,%ecx
  setne  %al
  add    $0x5,%eax
  ret
```

assistant: ```c
int t1_init_ctxno_zc(int f, int orient) {
  int h, v, d, n, t, hv;
  n = 0;
  h = ((f & 0x0080) != 0) + ((f & 0x0020) != 0);
  v = ((f & 0x0010) != 0) + ((f & 0x0040) != 0);
  d = ((f & 0x0008) != 0) + ((f & 0x0001) != 0) + ((f & 0x0002) != 0) + ((f & 0x0004) != 0);
  switch (orient) {
  case 2:
    t = h;
    h = v;
    v = t;
  case 0:
  case 1:
    if (!h) {
      if (!v) {
        if (!d)
          n = 0;
        else if (d == 1)
          n = 1;
        else
          n = 2;
      } else if (v == 1) {
        n = 3;
      } else {
        n = 4;
      }
    } else if (h == 1) {
      if (!v) {
        if (!d)
          n = 5;
        else
          n = 6;
      } else {
        n = 7;
      }
    } else
      n = 8;
    break;
  case 3:
    hv = h + v;
    if (!d) {
      if (!hv) {
        n = 0;
      } else if (hv == 1) {
        n = 1;
      } else {
        n = 2;
      }
    } else if (d == 1) {
      if (!hv) {
        n = 3;
      } else if (hv == 1) {
        n = 4;
      } else {
        n = 5;
      }
    } else if (d == 2) {
      if (!hv) {
        n = 6;
      } else {
        n = 7;
      }
    } else {
      n = 8;
    }
    break;
  }
  return ((0 + 1) + n);
}
```