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Bitboards and pinned pieces

As I mentioned in another thread, I wrote a little magic number generator yesterday. Because it would be a waste to produce these numbers without using them for anything, I've now modified the bitboard version of my program (which used classical rotated bitboards) to use magic-based attack generation instead.

Most of the work was very easy, but I had a few small problems. A problem I still haven't been able to solve cleanly is to find a bitboard of all pinned pieces for a given side. My old rotated bitboard code looks like this:
Code:

Bitboard Position::pinned_pieces(Color c) const {
  Bitboard pinned = EmptyBoardBB, b1, b2;
  Square ksq = this->king_square(c), s;
  Color them = opposite_color(c);
  
  for(Direction d = DIR_E; d <= DIR_NW; d++) {
    b1 = this->sliders_of_color(them, d);
    b2 = this->sliding_attacks(ksq, d) & this->pieces_of_color(c);
    while(b2) {
      s = pop_1st_bit(&b2);
      if(this->sliding_attacks(s, d) & b1) set_bit(&pinned, s);
    }
  }
  return pinned;
}

My current magic bitboard version is a lot messier:
Code:

Bitboard Position::pinned_pieces(Color c) const {
  Bitboard b1, b2, b3, pinned, sliders;
  Square ksq = this->king_square(c);
  Color them = opposite_color(c);

  pinned = EmptyBoardBB;
  b1 = this->occupied_squares();

  sliders = this->rooks_and_queens_of_color(them);
  if(sliders) {
    b2 = this->rook_attacks(ksq) & this->pieces_of_color(c);
    while(b2) {
      b3 = b2 & -b2;
      b2 ^= b3;
      if(rook_attacks_bb(ksq, b1 ^ b3) & sliders) pinned |= b3;
    }
  }
  sliders = this->bishops_and_queens_of_color(them);
  if(sliders) {
    b2 = this->bishop_attacks(ksq) & this->pieces_of_color(c);
    while(b2) {
      b3 = b2 & -b2;
      b2 ^= b3;
      if(bishop_attacks_bb(ksq, b1 ^ b3) & sliders) pinned |= b3;
    }
  }
  return pinned;
}

This is fast enough and has the advantage of avoiding the dreaded bitscanning, but looks awfully ugly and complicated. There has to be a better way to do this. Because magic bitboards generate attacks in several directions at once, I would expect a good magic bitboard solution to be shorter and simpler than the rotated bitboard solution.

Any ideas about how to improve this code?

Tord

位棋盘可用来做评估，但中象是９Ｘ１０格的，怎样在中象中解决这个问题呢，奇兵很好的解决了这个问题，将棋盘分成上下二部分，这样就能很好的用ＭＭＸ的６４位指令来优化了．

ＰＩＮ，　主要用来做解将算法，但中象多了一个炮，与马的别脚，就复杂多了．