user: 提供的汇编代码用C语言表示会是什么样子：

```asm
<convert2to8>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  sub    $0x40,%rsp
  mov    %rdi,-0x28(%rbp)
  mov    %rsi,-0x30(%rbp)
  mov    %edx,-0x34(%rbp)
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  movl   $0xf6a35c09,-0xc(%rbp)
  mov    D21(%rip),%eax
  test   %eax,%eax
  je     L14
  movl   $0x0,-0x1c(%rbp)
  jmp    L2
L41:
  movl   $0x0,-0x14(%rbp)
  jmp    L58
L1:
  mov    -0x14(%rbp),%eax
  add    %eax,%eax
  mov    -0x1c(%rbp),%edx
  mov    %eax,%ecx
  sar    %cl,%edx
  mov    %edx,%eax
  and    $0x3,%eax
  mov    %eax,-0x10(%rbp)
  mov    -0x10(%rbp),%eax
  cltq
  movzbl -0xc(%rbp,%rax,1),%eax
  mov    -0x14(%rbp),%edx
  movslq %edx,%rcx
  mov    -0x1c(%rbp),%edx
  movslq %edx,%rdx
  shl    $0x2,%rdx
  add    %rdx,%rcx
  lea    D20(%rip),%rdx
  add    %rcx,%rdx
  mov    %al,(%rdx)
  addl   $0x1,-0x14(%rbp)
L58:
  cmpl   $0x3,-0x14(%rbp)
  jle    L1
  addl   $0x1,-0x1c(%rbp)
L2:
  cmpl   $0xff,-0x1c(%rbp)
  jle    L41
L14:
  movl   $0x0,-0x18(%rbp)
  movl   $0x0,-0x1c(%rbp)
  jmp    L34
L19:
  mov    -0x1c(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movzbl %al,%ecx
  mov    -0x18(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x18(%rbp)
  movslq %eax,%rdx
  mov    -0x28(%rbp),%rax
  add    %rax,%rdx
  movslq %ecx,%rax
  lea    0x0(,%rax,4),%rcx
  lea    D20(%rip),%rax
  movzbl (%rcx,%rax,1),%eax
  mov    %al,(%rdx)
  mov    -0x1c(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movzbl %al,%ecx
  mov    -0x18(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x18(%rbp)
  movslq %eax,%rdx
  mov    -0x28(%rbp),%rax
  add    %rax,%rdx
  movslq %ecx,%rax
  lea    0x0(,%rax,4),%rcx
  lea    D52(%rip),%rax
  movzbl (%rcx,%rax,1),%eax
  mov    %al,(%rdx)
  mov    -0x1c(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movzbl %al,%ecx
  mov    -0x18(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x18(%rbp)
  movslq %eax,%rdx
  mov    -0x28(%rbp),%rax
  add    %rax,%rdx
  movslq %ecx,%rax
  lea    0x0(,%rax,4),%rcx
  lea    D44(%rip),%rax
  movzbl (%rcx,%rax,1),%eax
  mov    %al,(%rdx)
  mov    -0x1c(%rbp),%eax
  movslq %eax,%rdx
  mov    -0x30(%rbp),%rax
  add    %rdx,%rax
  movzbl (%rax),%eax
  movzbl %al,%ecx
  mov    -0x18(%rbp),%eax
  lea    0x1(%rax),%edx
  mov    %edx,-0x18(%rbp)
  movslq %eax,%rdx
  mov    -0x28(%rbp),%rax
  add    %rax,%rdx
  movslq %ecx,%rax
  lea    0x0(,%rax,4),%rcx
  lea    D25(%rip),%rax
  movzbl (%rcx,%rax,1),%eax
  mov    %al,(%rdx)
  addl   $0x1,-0x1c(%rbp)
L34:
  mov    -0x1c(%rbp),%eax
  cmp    -0x34(%rbp),%eax
  jl     L19
  nop
  mov    -0x8(%rbp),%rax
  sub    %fs:0x28,%rax
  je     L67
  call   <__stack_chk_fail@plt>
L67:
  leave
  ret
```

assistant: ```c
void convert2to8(unsigned char *dest, const unsigned char *src, int bytes) {
  const unsigned char levels[4] = {9, 92, 163, 246};
  static int first = 1;
  static unsigned char lut2to8[256][4];
  int i, o;
  if (first) {
    for (i = 0; i < 256; ++i) {
      int j;
      for (j = 0; j < 4; ++j) {
        int k;
        k = (i >> (2 * j)) & 0x3;
        lut2to8[i][j] = levels[k];
      }
    }
  }
  o = 0;
  for (i = 0; i < bytes; ++i) {
    dest[o++] = lut2to8[src[i]][0];
    dest[o++] = lut2to8[src[i]][1];
    dest[o++] = lut2to8[src[i]][2];
    dest[o++] = lut2to8[src[i]][3];
  }
}
```