問題匯出

move i = do
    x <- left i
    tell "Go"
    y <- left x
    return y

其中

left i = writer ( x -1,"move left\t')
:t tell "G0" :: MonadWriter [Char] m => m ()

問題

tell "Go"的上下文到底在哪裡

我們一步步的解開

move  :: Int -> Writer String Int
move i = do
    x <- left' i
    _ <- tell "GO\t"
 

    return x

我們把do展開

move i = let
            ok p = do
                    tell "go"
                    return p
            ok _ = fail ".."
           in writer ((i-1),"Move Left\t") >>= ok

展開tell

move' i = let
            ok p = do
                    writer ((),"Go\t")
                    return
 
 p
            ok _ = fail ".."
           in writer ((i-1),"Move Left\t") >>= ok

將let解構轉換為where解構,暫時消除掉i

move i = writer ((i-1),"Move Left\t") >>= ok where
            ok p = do
                    writer ((),"Go\t")
                    return p

消除掉第二個do

move i = writer ((i-1),"Move Left\t"
 
) >>= ok where
            ok p = writer ((),"Go\t") >>  return p

增加輔助函式，做最終轉換

instance (Monoid w) => Monad (Writer w) where
    return x = Writer (x,mempty)
    (Writer (x,v)) >>= f = let (Writer ( y,v')) = fx
        in Writer (y, v `mappend` v')

temp ::Int -> Writer String Int
temp i = writer ((),"GO\t") >> return (i-1)

move i =  case runWriter(temp i) of
           (y,v') -> writer (y, "Move Left\t" `mappend` v')
           _ -> writer(0,mempty)

對於temp的模擬，可以參看monad這裡的對於>>的實現，確實是使用了>>=。這樣既滿足了需要序列運算上下文，又能保證丟棄了上文的M a的a，但是仍然可以把M傳遞給下文。而恰好，Writer的宣告形式是Writer String a,這樣可以把a丟棄掉，而把String傳遞給下文，達到了tell要求的僅保留String內容，而忽略a的內容.
結論：
可以看出，上下文首先是因為做了內部的do運算，得到的關鍵的Writer((),"Go")，
之後把(i-1) 附屬到了tell的結果上，然後再做外部運算，結束掉x <- left i。