user: 这段汇编代码用C语言表示是什么样的?

```asm
<idctrow_enc>:
  endbr64
  mov    %rdi,%rax
  movswl 0x8(%rdi),%r9d
  shl    $0xb,%r9d
  movzwl 0xc(%rdi),%ecx
  movzwl 0x4(%rdi),%r10d
  movzwl 0x2(%rdi),%r8d
  movzwl 0xe(%rdi),%edi
  movzwl 0xa(%rax),%r11d
  movzwl 0x6(%rax),%esi
  mov    %ecx,%edx
  or     %r10d,%edx
  or     %r8d,%edx
  or     %edi,%edx
  or     %r11d,%edx
  or     %esi,%edx
  movswl %dx,%edx
  or     %r9d,%edx
  je     L2
  push   %rbp
  push   %rbx
  movswl %cx,%ecx
  movswl %r10w,%r10d
  movswl %r8w,%edx
  movswl %di,%edi
  movswl %r11w,%r11d
  movswl %si,%esi
  movswl (%rax),%r8d
  shl    $0xb,%r8d
  sub    $0xffffff80,%r8d
  lea    (%rdx,%rdi,1),%ebx
  imul   $0x235,%ebx,%ebx
  imul   $0x8e4,%edx,%edx
  add    %ebx,%edx
  imul   $0xfffff2b2,%edi,%edi
  add    %ebx,%edi
  lea    (%r11,%rsi,1),%ebx
  imul   $0x968,%ebx,%ebx
  imul   $0xfffffce1,%r11d,%r11d
  add    %ebx,%r11d
  imul   $0xfffff04f,%esi,%esi
  add    %ebx,%esi
  lea    (%r9,%r8,1),%ebx
  sub    %r9d,%r8d
  lea    (%rcx,%r10,1),%ebp
  imul   $0x454,%ebp,%ebp
  imul   $0xfffff138,%ecx,%ecx
  add    %ebp,%ecx
  imul   $0x620,%r10d,%r9d
  add    %ebp,%r9d
  lea    (%rdx,%r11,1),%r10d
  sub    %r11d,%edx
  lea    (%rdi,%rsi,1),%r11d
  sub    %esi,%edi
  lea    (%rbx,%r9,1),%esi
  sub    %r9d,%ebx
  lea    (%r8,%rcx,1),%r9d
  sub    %ecx,%r8d
  lea    (%rdx,%rdi,1),%ecx
  imul   $0xb5,%ecx,%ecx
  sub    $0xffffff80,%ecx
  sar    $0x8,%ecx
  sub    %edi,%edx
  imul   $0xb5,%edx,%edx
  sub    $0xffffff80,%edx
  sar    $0x8,%edx
  lea    (%r10,%rsi,1),%edi
  sar    $0x8,%edi
  mov    %di,(%rax)
  lea    (%r9,%rcx,1),%edi
  sar    $0x8,%edi
  mov    %di,0x2(%rax)
  lea    (%r8,%rdx,1),%edi
  sar    $0x8,%edi
  mov    %di,0x4(%rax)
  lea    (%r11,%rbx,1),%edi
  sar    $0x8,%edi
  mov    %di,0x6(%rax)
  sub    %r11d,%ebx
  sar    $0x8,%ebx
  mov    %bx,0x8(%rax)
  sub    %edx,%r8d
  sar    $0x8,%r8d
  mov    %r8w,0xa(%rax)
  sub    %ecx,%r9d
  sar    $0x8,%r9d
  mov    %r9w,0xc(%rax)
  sub    %r10d,%esi
  sar    $0x8,%esi
  mov    %si,0xe(%rax)
  pop    %rbx
  pop    %rbp
  ret
L2:
  movswl (%rax),%edx
  shl    $0x3,%edx
  mov    %dx,0xe(%rax)
  mov    %dx,0xc(%rax)
  mov    %dx,0xa(%rax)
  mov    %dx,0x8(%rax)
  mov    %dx,0x6(%rax)
  mov    %dx,0x4(%rax)
  mov    %dx,0x2(%rax)
  mov    %dx,(%rax)
  ret
```

assistant: ```c
void idctrow_enc(blk) short *blk;
{
  int x0, x1, x2, x3, x4, x5, x6, x7, x8;
  if (!((x1 = blk[4] << 11) | (x2 = blk[6]) | (x3 = blk[2]) | (x4 = blk[1]) | (x5 = blk[7]) | (x6 = blk[5]) | (x7 = blk[3]))) {
    blk[0] = blk[1] = blk[2] = blk[3] = blk[4] = blk[5] = blk[6] = blk[7] = blk[0] << 3;
    return;
  }
  x0 = (blk[0] << 11) + 128;
  x8 = 565 * (x4 + x5);
  x4 = x8 + (2841 - 565) * x4;
  x5 = x8 - (2841 + 565) * x5;
  x8 = 2408 * (x6 + x7);
  x6 = x8 - (2408 - 1609) * x6;
  x7 = x8 - (2408 + 1609) * x7;
  x8 = x0 + x1;
  x0 -= x1;
  x1 = 1108 * (x3 + x2);
  x2 = x1 - (2676 + 1108) * x2;
  x3 = x1 + (2676 - 1108) * x3;
  x1 = x4 + x6;
  x4 -= x6;
  x6 = x5 + x7;
  x5 -= x7;
  x7 = x8 + x3;
  x8 -= x3;
  x3 = x0 + x2;
  x0 -= x2;
  x2 = (181 * (x4 + x5) + 128) >> 8;
  x4 = (181 * (x4 - x5) + 128) >> 8;
  blk[0] = (x7 + x1) >> 8;
  blk[1] = (x3 + x2) >> 8;
  blk[2] = (x0 + x4) >> 8;
  blk[3] = (x8 + x6) >> 8;
  blk[4] = (x8 - x6) >> 8;
  blk[5] = (x0 - x4) >> 8;
  blk[6] = (x3 - x2) >> 8;
  blk[7] = (x7 - x1) >> 8;
}
```