user: 这段汇编代码用C语言表示是什么样的？

```asm
<fFunction>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0xe0,%rsp
  mov    %edi,%edx
  mov    %esi,%eax
  mov    %dx,-0xd4(%rbp)
  mov    %ax,-0xd8(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0x1,-0x90(%rbp)
  movl   $0x0,-0x8c(%rbp)
  movl   $0x3,-0x88(%rbp)
  movl   $0x2,-0x84(%rbp)
  movl   $0x3,-0x80(%rbp)
  movl   $0x2,-0x7c(%rbp)
  movl   $0x1,-0x78(%rbp)
  movl   $0x0,-0x74(%rbp)
  movl   $0x0,-0x70(%rbp)
  movl   $0x2,-0x6c(%rbp)
  movl   $0x1,-0x68(%rbp)
  movl   $0x3,-0x64(%rbp)
  movl   $0x3,-0x60(%rbp)
  movl   $0x1,-0x5c(%rbp)
  movl   $0x3,-0x58(%rbp)
  movl   $0x2,-0x54(%rbp)
  movl   $0x0,-0x50(%rbp)
  movl   $0x1,-0x4c(%rbp)
  movl   $0x2,-0x48(%rbp)
  movl   $0x3,-0x44(%rbp)
  movl   $0x2,-0x40(%rbp)
  movl   $0x0,-0x3c(%rbp)
  movl   $0x1,-0x38(%rbp)
  movl   $0x3,-0x34(%rbp)
  movl   $0x3,-0x30(%rbp)
  movl   $0x0,-0x2c(%rbp)
  movl   $0x1,-0x28(%rbp)
  movl   $0x0,-0x24(%rbp)
  movl   $0x2,-0x20(%rbp)
  movl   $0x1,-0x1c(%rbp)
  movl   $0x0,-0x18(%rbp)
  movl   $0x3,-0x14(%rbp)
  movl   $0x4,-0xb0(%rbp)
  movl   $0x1,-0xac(%rbp)
  movl   $0x2,-0xa8(%rbp)
  movl   $0x3,-0xa4(%rbp)
  movl   $0x2,-0xa0(%rbp)
  movl   $0x3,-0x9c(%rbp)
  movl   $0x4,-0x98(%rbp)
  movl   $0x1,-0x94(%rbp)
  movl   $0x2,-0xc0(%rbp)
  movl   $0x4,-0xbc(%rbp)
  movl   $0x3,-0xb8(%rbp)
  movl   $0x1,-0xb4(%rbp)
  movzwl -0xd4(%rbp),%eax
  lea    -0xb0(%rbp),%rsi
  mov    $0x8,%ecx
  mov    $0x4,%edx
  mov    %eax,%edi
  call   <permute@plt>
  mov    %ax,-0xc2(%rbp)
  movzwl -0xd8(%rbp),%eax
  xor    %ax,-0xc2(%rbp)
  movzwl -0xc2(%rbp),%eax
  sar    $0x6,%eax
  and    $0x2,%eax
  mov    %eax,%edx
  movzwl -0xc2(%rbp),%eax
  sar    $0x4,%eax
  and    $0x1,%eax
  or     %eax,%edx
  movzwl -0xc2(%rbp),%eax
  sar    $0x5,%eax
  and    $0x3,%eax
  movslq %eax,%rcx
  movslq %edx,%rax
  shl    $0x2,%rax
  add    %rcx,%rax
  mov    -0x90(%rbp,%rax,4),%eax
  lea    0x0(,%rax,4),%esi
  movzwl -0xc2(%rbp),%eax
  sar    $0x2,%eax
  and    $0x2,%eax
  mov    %eax,%edx
  movzwl -0xc2(%rbp),%eax
  and    $0x1,%eax
  or     %eax,%edx
  movzwl -0xc2(%rbp),%eax
  sar    %eax
  and    $0x3,%eax
  movslq %eax,%rcx
  movslq %edx,%rax
  shl    $0x2,%rax
  add    %rcx,%rax
  mov    -0x50(%rbp,%rax,4),%eax
  add    %esi,%eax
  mov    %ax,-0xc2(%rbp)
  movzwl -0xc2(%rbp),%eax
  lea    -0xc0(%rbp),%rsi
  mov    $0x4,%ecx
  mov    $0x4,%edx
  mov    %eax,%edi
  call   <permute@plt>
  mov    %ax,-0xc2(%rbp)
  movzwl -0xc2(%rbp),%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L8
  call   <__stack_chk_fail@plt>
L8:
  leave
  ret
```

assistant: ```c
unsigned short fFunction(unsigned short r, unsigned short key) {
  int sZero[4][4] = {{1, 0, 3, 2}, {3, 2, 1, 0}, {0, 2, 1, 3}, {3, 1, 3, 2}};
  int sOne[4][4] = {{0, 1, 2, 3}, {2, 0, 1, 3}, {3, 0, 1, 0}, {2, 1, 0, 3}};
  int eP[8] = {4, 1, 2, 3, 2, 3, 4, 1};
  int pFour[4] = {2, 4, 3, 1};
  unsigned short expR;
  expR = permute(r, eP, 4, 8);
  expR = expR ^ key;
  expR = (sZero[((expR & 128) >> 6) + ((expR & 16) >> 4)][((expR & 64) >> 5) + ((expR & 32) >> 5)] << 2) + sOne[((expR & 8) >> 2) + (expR & 1)][((expR & 4) >> 1) + ((expR & 2) >> 1)];
  expR = permute(expR, pFour, 4, 4);
  return expR;
}
```