如何同时返回已处理的列表和修改的对象？

问：

我正在设计一个名为 Daphne 的密码，它采用一个明文字节并返回一个加密字节和一个修改后的 Daphne。当应用于列表时，它应该在处理完所有字节后返回加密字节列表和 Daphne。我不确定如何在不占用 O（n） 内存的情况下执行此操作。代码如下：

byteEncrypt :: Daphne -> Word8 -> (Daphne,Word8)
byteEncrypt (Daphne key sreg acc) plain = ((Daphne key newsreg newacc),crypt) where
  left = computeLeft key sreg acc
  right = computeRight key sreg acc
  crypt = step plain left right
  newacc = acc+plain
  newsreg = Seq.drop 1 (sreg |> crypt)

byteDecrypt :: Daphne -> Word8 -> (Daphne,Word8)
byteDecrypt (Daphne key sreg acc) crypt = ((Daphne key newsreg newacc),plain) where
  left = computeLeft key sreg acc
  right = computeRight key sreg acc
  plain = invStep crypt left right
  newacc = acc+plain
  newsreg = Seq.drop 1 (sreg |> crypt)

seqEncrypt :: Seq.Seq Word8 -> (Daphne,Seq.Seq Word8) -> (Daphne,Seq.Seq Word8)
seqEncrypt Seq.Empty a = a
seqEncrypt (bs:|>b) (daph,acc) = (daph2,acc2) where
  (daph1,acc1) = seqEncrypt bs (daph,acc)
  (daph2,c) = byteEncrypt daph1 b
  acc2 = acc1 |> c

listEncrypt :: Daphne -> [Word8] -> (Daphne,[Word8])
listEncrypt daph bs = (daph1,toList seq1) where
  (daph1,seq1) = seqEncrypt (Seq.fromList bs) (daph,Seq.Empty)

seqDecrypt :: Seq.Seq Word8 -> (Daphne,Seq.Seq Word8) -> (Daphne,Seq.Seq Word8)
seqDecrypt Seq.Empty a = a
seqDecrypt (bs:|>b) (daph,acc) = (daph2,acc2) where
  (daph1,acc1) = seqDecrypt bs (daph,acc)
  (daph2,c) = byteDecrypt daph1 b
  acc2 = acc1 |> c

listDecrypt :: Daphne -> [Word8] -> (Daphne,[Word8])
listDecrypt daph bs = (daph1,toList seq1) where
  (daph1,seq1) = seqDecrypt (Seq.fromList bs) (daph,Seq.Empty)

listEncrypt在RAM中有一个列表的副本作为序列，它占用了O（n）个RAM，这应该不是必需的。

完整代码位于 https://github.com/phma/daphne 。它还没有准备好投入生产;我没有对它进行过加密分析，更不用说其他人了。我可能会通过它运行千兆字节的数据并计算统计数据。