user: 这段汇编代码用C语言表示是什么样的？

```asm
<_shiftR>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %edx,%eax
  mov    %al,-0x34(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movzbl -0x34(%rbp),%eax
  shr    $0x3,%al
  movzbl %al,%eax
  mov    %eax,-0x18(%rbp)
  movzbl -0x34(%rbp),%eax
  and    $0x7,%eax
  mov    %eax,-0x14(%rbp)
  movl   $0x0,-0x1c(%rbp)
  jmp    L0
L74:
  mov    -0x1c(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movb   $0x0,(%rax)
  addl   $0x1,-0x1c(%rbp)
L0:
  mov    -0x1c(%rbp),%eax
  cmp    -0x18(%rbp),%eax
  jl     L74
  mov    -0x18(%rbp),%eax
  mov    %eax,-0x1c(%rbp)
  jmp    L108
L98:
  mov    -0x1c(%rbp),%eax
  sub    -0x18(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  mov    -0x1c(%rbp),%edx
  movslq %edx,%rcx
  mov    -0x28(%rbp),%rdx
  add    %rcx,%rdx
  movzbl (%rax),%eax
  mov    %al,(%rdx)
  addl   $0x1,-0x1c(%rbp)
L108:
  cmpl   $0x7,-0x1c(%rbp)
  jle    L98
  movl   $0x0,-0x1c(%rbp)
  jmp    L80
L65:
  mov    -0x1c(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%edx
  mov    -0x1c(%rbp),%eax
  cltq
  mov    %dl,-0x10(%rbp,%rax,1)
  addl   $0x1,-0x1c(%rbp)
L80:
  cmpl   $0x7,-0x1c(%rbp)
  jle    L65
  cmpl   $0x0,-0x14(%rbp)
  jle    L77
  mov    -0x18(%rbp),%eax
  mov    %eax,-0x1c(%rbp)
  jmp    L21
L15:
  mov    -0x1c(%rbp),%eax
  cmp    -0x18(%rbp),%eax
  jne    L5
  mov    -0x1c(%rbp),%eax
  cltq
  movzbl -0x10(%rbp,%rax,1),%eax
  movzbl %al,%edx
  mov    -0x14(%rbp),%eax
  mov    %eax,%ecx
  sar    %cl,%edx
  mov    %edx,%ecx
  mov    -0x1c(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    %ecx,%edx
  mov    %dl,(%rax)
  jmp    L73
L5:
  mov    -0x1c(%rbp),%eax
  cltq
  movzbl -0x10(%rbp,%rax,1),%eax
  movzbl %al,%edx
  mov    -0x14(%rbp),%eax
  mov    %eax,%ecx
  sar    %cl,%edx
  mov    %edx,%eax
  mov    %eax,%esi
  mov    -0x1c(%rbp),%eax
  sub    $0x1,%eax
  cltq
  movzbl -0x10(%rbp,%rax,1),%eax
  movzbl %al,%edx
  mov    $0x8,%eax
  sub    -0x14(%rbp),%eax
  mov    %eax,%ecx
  shl    %cl,%edx
  mov    %edx,%eax
  or     %eax,%esi
  mov    %esi,%ecx
  mov    -0x1c(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x28(%rbp),%rax
  add    %rdx,%rax
  mov    %ecx,%edx
  mov    %dl,(%rax)
L73:
  addl   $0x1,-0x1c(%rbp)
L21:
  cmpl   $0x7,-0x1c(%rbp)
  jle    L15
L77:
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L19
  call   <__stack_chk_fail@plt>
L19:
  leave
  ret
```

assistant: ```c
void _shiftR(unsigned char ans[8], unsigned char x[8], unsigned char r) {
  int i;
  int shiftbytes = r / 8;
  int shiftbits = r % 8;
  unsigned char temp[8];
  for (i = 0; i < shiftbytes; i++) {
    ans[i] = 0;
  }
  for (i = shiftbytes; i < 8; i++) {
    ans[i] = x[i - shiftbytes];
  }
  for (i = 0; i < 8; i++) {
    temp[i] = ans[i];
  }
  if (shiftbits > 0) {
    for (i = shiftbytes; i < 8; i++) {
      if (i == shiftbytes) {
        ans[i] = (temp[i] >> shiftbits);
      } else {
        ans[i] = (temp[i] >> shiftbits) | (temp[i - 1] << (8 - shiftbits));
      }
    }
  }
}
```