Question

有人可以向我解释为什么要说教＆＃34;％broadcast.splatinsert.i = insertelement＆lt; 4 x i32＆gt; undef，i32％reverse.idx.i，i32 0＆＃34;
打印＆＃34;指令尚不可解释！？

lli ver 3.3 完整来源https://www.sendspace.com/file/e9kgng

代码块：

vector.body.i:                                    ; preds = %vector.body.i, %for.body.lr.ph.i
  %index.i = phi i64 [ %index.next.i, %vector.body.i ], [ 0, %for.body.lr.ph.i ]
  %vec.phi.i = phi <4 x i32> [ %86, %vector.body.i ], [ <i32 1, i32 1, i32 1, i32 1>, %for.body.lr.ph.i ]
  %vec.phi11.i = phi <4 x i32> [ %87, %vector.body.i ], [ <i32 1, i32 1, i32 1, i32 1>, %for.body.lr.ph.i ]
  %resize.norm.idx.i = trunc i64 %index.i to i32
  %reverse.idx.i = sub i32 %.x.i113, %resize.norm.idx.i
  %broadcast.splatinsert.i = insertelement <4 x i32> undef, i32 %reverse.idx.i, i32 0
  %broadcast.splat.i = shufflevector <4 x i32> %broadcast.splatinsert.i, <4 x i32> undef, <4 x i32> zeroinitializer
  %induction.i = add <4 x i32> %broadcast.splat.i, <i32 0, i32 -1, i32 -2, i32 -3>
  %84 = load i64* %main_HideLocalConstant_variable_14
  %main_hide_const_value83 = xor i64 %84, 8
  %induction12.i = add <4 x i32> %broadcast.splat.i, <i32 -4, i32 -5, i32 -6, i32 -7>
  %85 = load i64* %main_HideLocalConstant_variable_4
  %main_hide_const_value21 = xor i64 %85, %main_hide_const_value83
  %86 = mul <4 x i32> %vec.phi.i, %induction.i
  %87 = mul <4 x i32> %vec.phi11.i, %induction12.i
  %index.next.i = add i64 %index.i, %main_hide_const_value21
  %88 = icmp eq i64 %index.next.i, %n.vec.i
  br i1 %88, label %middle.block.i, label %vector.body.i

Answer 1

有一个＆＃34; undef＆＃34;你看到的操作数，因此无法解释指令。如果这个bitcode是lvv 3.3中LoopVectorize传递的输出，那么我会看到它可能是一个可能已在新版本中修复的bug。

lli指令尚未解释

1 个答案: