这是这个问题的后续行动.我正在尝试将@ErikR的shell的回答合并到我的InputT循环中。
main :: IO [String]
main = do
c <- makeCounter
execStateT (repl c) []
repl :: Counter -> StateT [String] IO ()
repl c = lift $ runInputT defaultSettings loop
where
loop = do
minput <- getLineIO $ in_ps1 $ c
case minput of
Nothing -> lift $ outputStrLn "Goodbye."
Just input -> (liftIO $ process c input) >> loop
getLineIO :: (MonadException m) => IO String -> InputT m (Maybe String)
getLineIO ios = do
s <- liftIO ios
getInputLine s并得到一个错误
Main.hs:59:10:
Couldn't match type ‘InputT m0’ with ‘IO’
Expected type: StateT [String] IO ()
Actual type: StateT [String] (InputT m0) ()
Relevant bindings include
loop :: InputT (InputT m0) () (bound at Main.hs:61:3)
In the expression: lift $ runInputT defaultSettings loop
In an equation for ‘repl’:
repl c
= lift $ runInputT defaultSettings loop
where
loop
= do { minput <- getLineIO $ in_ps1 $ c;
.... }
Main.hs:62:5:
No instance for (Monad m0) arising from a do statement
The type variable ‘m0’ is ambiguous
Relevant bindings include
loop :: InputT (InputT m0) () (bound at Main.hs:61:3)
Note: there are several potential instances:
instance Monad (Text.Parsec.Prim.ParsecT s u m)
-- Defined in ‘Text.Parsec.Prim’
instance Monad (Either e) -- Defined in ‘Data.Either’
instance Monad Data.Proxy.Proxy -- Defined in ‘Data.Proxy’
...plus 15 others
In a stmt of a 'do' block: minput <- getLineIO $ in_ps1 $ c
In the expression:
do { minput <- getLineIO $ in_ps1 $ c;
case minput of {
Nothing -> lift $ outputStrLn "Goodbye."
Just input -> (liftIO $ process c input) >> loop } }
In an equation for ‘loop’:
loop
= do { minput <- getLineIO $ in_ps1 $ c;
case minput of {
Nothing -> lift $ outputStrLn "Goodbye."
Just input -> (liftIO $ process c input) >> loop } }我知道haskelline对历史有一个内置的支持,但我正在尝试将其作为一个练习来实现。
请随时建议更换单台变压器,以获得相同的功能。
我真正的问题
我想在写一个Haskell中将类似于ipython的功能添加到lambda中,即:
I.输入和输出计数器,将出现在提示符中,即
In[1]>
Out[1]>这已经是完成了。
将每个命令保存到历史记录中(自动),并使用一个特殊命令显示所有以前的命令,例如histInput (与ipython中的hist相同)。另外,保存所有输出结果的历史记录,并使用histOutput显示它们。这就是我在这个问题上要做的(目前只输入历史)。
引用以前的输入和输出,例如,如果In[1]是x,那么In[1] + 2应该由x + 2代替,输出也应该替换。
更新
我尝试将@ErikR的回答和临时禁用的showStep结合起来,得到如下结果:
module Main where
import Syntax
import Parser
import Eval
import Pretty
import Counter
import Control.Monad
import Control.Monad.Trans
import System.Console.Haskeline
import Control.Monad.State
showStep :: (Int, Expr) -> IO ()
showStep (d, x) = putStrLn ((replicate d ' ') ++ "=> " ++ ppexpr x)
process :: Counter -> String -> InputT (StateT [String] IO) ()
process c line =
if ((length line) > 0)
then
if (head line) /= '%'
then do
modify (++ [line])
let res = parseExpr line
case res of
Left err -> outputStrLn $ show err
Right ex -> do
let (out, ~steps) = runEval ex
--mapM_ showStep steps
out_ps1 c $ out2iout $ show out
else do
let iout = handle_cmd line
out_ps1 c iout
-- TODO: don't increment counter for empty lines
else do
outputStrLn ""
out2iout :: String -> IO String
out2iout s = return s
out_ps1 :: Counter -> IO String -> InputT (StateT [String] IO) ()
out_ps1 c iout = do
out <- liftIO iout
let out_count = c 0
outputStrLn $ "Out[" ++ (show out_count) ++ "]: " ++ out
outputStrLn ""
handle_cmd :: String -> IO String
handle_cmd line = if line == "%hist"
then
evalStateT getHist []
else
return "unknown cmd"
getHist :: StateT [String] IO String
getHist = do
hist <- lift get
forM_ (zip [(1::Int)..] hist) $ \(i, h) -> do
show i ++ ": " ++ show h
main :: IO ()
main = do
c <- makeCounter
repl c
repl :: Counter -> IO ()
repl c = evalStateT (runInputT defaultSettings(loop c)) []
loop :: Counter -> InputT (StateT [String] IO) ()
loop c = do
minput <- getLineIO $ in_ps1 $ c
case minput of
Nothing -> return ()
Just input -> process c input >> loop c
getLineIO :: (MonadException m) => IO String -> InputT m (Maybe String)
getLineIO ios = do
s <- liftIO ios
getInputLine s
in_ps1 :: Counter -> IO String
in_ps1 c = do
let ion = c 1
n <- ion
let s = "Untyped: In[" ++ (show n) ++ "]> "
return s它仍然没有编译:
Main.hs:59:5:
Couldn't match type ‘[]’ with ‘StateT [String] IO’
Expected type: StateT [String] IO String
Actual type: [()]
In a stmt of a 'do' block:
forM_ (zip [(1 :: Int) .. ] hist)
$ \ (i, h) -> do { show i ++ ": " ++ show h }
In the expression:
do { hist <- lift get;
forM_ (zip [(1 :: Int) .. ] hist) $ \ (i, h) -> do { ... } }
In an equation for ‘getHist’:
getHist
= do { hist <- lift get;
forM_ (zip [(1 :: Int) .. ] hist) $ \ (i, h) -> ... }发布于 2016-06-20 19:35:36
第一个错误是因为您已经声明了
main :: IO ()但同时也
execStateT (...) :: IO [String]execStateT返回计算的最终状态,您的状态为[String]类型。通常,这是通过不为main声明一个类型并让它被推断为某些a的IO a来解决的。第二个我不确定,但也许是一样的。
发布于 2016-06-23 00:03:58
我要猜猜你想做什么。
此程序识别以下命令:
hist -- show current history
add xxx -- add xxx to the history list
clear -- clear the history list
count -- show the count of history items
quit -- quit the command loop节目来源:
import System.Console.Haskeline
import Control.Monad.Trans.Class
import Control.Monad.Trans.State.Strict
import Control.Monad
main :: IO ()
main = evalStateT (runInputT defaultSettings loop) []
loop :: InputT (StateT [String] IO) ()
loop = do
minput <- getInputLine "% "
case minput of
Nothing -> return ()
Just "quit" -> return ()
Just input -> process input >> loop
process input = do
let args = words input
case args of
[] -> return ()
("hist": _) -> showHistory
("add" : x : _) -> lift $ modify (++ [x])
("clear": _) -> lift $ modify (const [])
("count": _) -> do hs <- lift get
outputStrLn $ "number of history items: " ++ show (length hs)
_ -> outputStrLn "???"
showHistory = do
hist <- lift get
forM_ (zip [(1::Int)..] hist) $ \(i,h) -> do
outputStrLn $ show i ++ " " ++ h发布于 2016-06-27 10:52:46
您拥有这里的代码编译,它将process定义为:
process :: Counter -> String -> IO ()若要创建具有此签名的process版本,请执行以下操作:
Counter -> String -> InputT (StateT [String] IO) ()只需使用liftIO:
process' :: Counter -> String -> InputT (StateT [String] IO) ()
process' counter str = liftIO $ process counter strhttps://stackoverflow.com/questions/37930204
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