是什么导致“Prelude.chr：错误的论点”？
What causes `Prelude.chr: bad argument`?
我编写了以下 Haskell 程序，其目的是像 Caesar cipher:
1 import System.IO
2 import System.Environment
3 import System.Exit
4 import Data.Char
5
6 shiftRight :: Int -> Char -> Char
7 shiftRight shift char = do
8 if isAsciiLower char
9 then if (toEnum (fromEnum char + shift) :: Char) > 'z'
10 then shiftRight (shift - 26) char
11 else toEnum (fromEnum char + shift) :: Char
12 else if isAsciiUpper char
13 then if (toEnum (fromEnum char + shift) :: Char) > 'Z'
14 then shiftRight (shift - 26) char
15 else toEnum (fromEnum char + shift) :: Char
16 else char
17
18 shiftLeft :: Int -> Char -> Char
19 shiftLeft shift char = do
20 if isAsciiLower char
21 then if (toEnum (fromEnum char - shift) :: Char) < 'a'
22 then shiftLeft (shift + 26) char
23 else toEnum (fromEnum char - shift) :: Char
24 else if isAsciiUpper char
25 then if (toEnum (fromEnum char - shift) :: Char) < 'A'
26 then shiftLeft (shift + 26) char
27 else toEnum (fromEnum char - shift) :: Char
28 else char
29
30 main = do
31 args <- getArgs
32 message <- getLine
33 case args of
34 [aString, aInt] ->
35 if aString == "-encode"
36 -- `read` converts aInt from string to int
37 -- `map` is used to apply `shiftRight` to each char in the string `message`
38 then putStrLn $ show $ map (shiftRight $ read $ aInt) message
39 else
40 if aString == "-decode"
41 then putStrLn $ show $ map (shiftLeft $ read $ aInt) message
42 else do
43 putStrLn ("Second argument should be either '-decode' or '-encode'!")
44 exitFailure
45 _ -> do
46 progName <- getProgName
47 putStrLn ("Usage: " ++ progName ++ " [-encode|-decode] [0-9]")
48 exitFailure
我的ghc版本如下：
$ ghc --version
The Glorious Glasgow Haskell Compilation System, version 8.6.5
我在 macos (Catalina) 上编译 Haskell:
$ ghc Prog1d.hs -o Prog1d
Loaded package environment from $HOME/.ghc/x86_64-darwin-8.6.5/environments/default
[1 of 1] Compiling Main ( Prog1d.hs, Prog1d.o )
Linking Prog1d ...
然后我运行我的代码：
$ echo "ABCXYZabcxyz" | ./Prog1d -encode 1
"BCDYZAbcdyza"
$ echo "ABCXYZabcxyz" | ./Prog1d -encode 2
"CDEZABcdezab"
$ echo "ABCXYZabcxyz" | ./Prog1d -encode 4
"EFGBCDefgbcd"
$ echo "ABCXYZabcxyz" | ./Prog1d -encode 100
"WXYTUVwxytuv"
$ echo "ABCXYZabcxyz" | ./Prog1d -decode 1
Prog1d: Prelude.chr: bad argument: (-14)
$ echo "ABCXYZabcxyz" | ./Prog1d -decode 2
Prog1d: Prelude.chr: bad argument: (-15)
$ echo "ABCXYZabcxyz" | ./Prog1d -decode 4
Prog1d: Prelude.chr: bad argument: (-17)
$ echo "ABCXYZabcxyz" | ./Prog1d -decode 100
Prog1d: Prelude.chr: bad argument: (-35)
为什么我得到 Prelude.chr: bad argument？是什么原因造成的，我该怎么做才能解决这个问题？
我读到过其他人遇到此错误，但在他们的情况下，删除 *.hi 文件解决了问题。我删除了 Prog1d.hi（以及 Prog1d.o 和 Prog1d），但没有任何效果。我觉得这可能是由我的代码中的某些内容引起的，可能是第 41 行：
then putStrLn $ show $ map (shiftLeft $ read $ aInt) message
但是这一行就像第 38 行一样，它适用于 -encode 用例。我一定遗漏了一些明显的东西。
我是 Haskell 的新手，所以请帮助我。
我主要习惯用C++、python、Java等命令式语言编写代码，对Haskell等函数式语言的思想和语法还不熟悉
感谢阅读本文！
我发现了问题。
在我的代码中，我将一个字符转换成它对应的ASCII码。
例如fromEnum 'A'会给你65。问题是，如果您将值从 ASCII 值的 table 移开，例如-1，你导致 toEnum 的错误参数错误，我用来比较以检查偏移是否过大。
例如：
echo "A" | ./Prog1d -decode 66
这将带我到程序中的第 25 行：
25 then if (toEnum (fromEnum char - shift) :: Char) < 'A'
fromEnum char 将是 fromEnum 'A' 这将给我 65.
然后我从 65 中减去 shift 值：66。所以，65 - 66 = -1.
然后 toEnum -1 导致 Prelude.chr: bad argument: (-1).
那是因为 -1 没有 ASCII 字符，该值超出范围。
这意味着我只需比较 int 值而不是 chars 来检查我的班次是否越界。
这是更正后的代码：
import System.IO
import System.Environment
import System.Exit
import Data.Char
shiftRight :: Int -> Char -> Char
shiftRight shift char = do
if isAsciiLower char
then if (fromEnum char + shift) > 122 -- 122 is 'z' in ASCII
then shiftRight (shift - 26) char
else toEnum (fromEnum char + shift) :: Char
else if isAsciiUpper char
then if (fromEnum char + shift) > 90 -- 90 is `Z` in ASCII
then shiftRight (shift - 26) char
else toEnum (fromEnum char + shift) :: Char
else char
shiftLeft :: Int -> Char -> Char
shiftLeft shift char = do
if isAsciiLower char
then if (fromEnum char - shift) < 97 -- 97 is 'a' in ASCII
then shiftLeft (shift - 26) char
else toEnum (fromEnum char - shift) :: Char
else if isAsciiUpper char
then if (fromEnum char - shift) < 65 -- 65 is 'A' in ASCII
then shiftLeft (shift - 26) char
else toEnum (fromEnum char - shift) :: Char
else char
main = do
args <- getArgs
message <- getLine
case args of
[aString, aInt] ->
if aString == "-encode"
-- `read` converts aInt from string to int
-- `map` is used to apply `shiftRight` to each char in the string `message`
then putStrLn $ map (shiftRight $ read $ aInt) message
else
if aString == "-decode"
then putStrLn $ map (shiftLeft $ read $ aInt) message
else do
putStrLn ("Second argument should be either '-decode' or '-encode'!")
exitFailure
_ -> do
progName <- getProgName
putStrLn ("Usage: " ++ progName ++ " [-encode|-decode] [0-9]")
exitFailure
这里是正确的、期望的输出：
$ echo "ABCXYZabcxyz" | ./Prog1d -encode 1
BCDYZAbcdyza
$ echo "ABCXYZabcxyz" | ./Prog1d -encode 2
CDEZABcdezab
$ echo "ABCXYZabcxyz" | ./Prog1d -encode 4
EFGBCDefgbcd
$ echo "ABCXYZabcxyz" | ./Prog1d -encode 100
WXYTUVwxytuv
$ echo "ABCXYZabcxyz" | ./Prog1d -decode 1
ZABWXYzabwxy
$ echo "ABCXYZabcxyz" | ./Prog1d -decode 2
YZAVWXyzavwx
$ echo "ABCXYZabcxyz" | ./Prog1d -decode 4
WXYTUVwxytuv
$ echo "ABCXYZabcxyz" | ./Prog1d -decode 100
EFGBCDefgbcd
编辑 (2020-09-01)：John Purdy 给了我一些反馈，这促使我从根本上重构我的代码。这是改进后的版本：
import System.IO
import System.Environment
import System.Exit
import Data.Char
shift :: Int -> Char -> Char
shift amount char
-- `ord` is `fromEnum` but only for `Char` types; converts a Char to an Int,
-- which gives us the ASCII number for that character
-- see: https://hackage.haskell.org/package/base-4.14.0.0/docs/Data-Char.html#v:ord
| isAsciiLower char && (shifted < ord 'a' || shifted > ord 'z') = shift cycled char
-- `chr` is `toEnum` but only for `Char` types; it converts from an Int to Char
-- see: https://hackage.haskell.org/package/base-4.14.0.0/docs/Data-Char.html#v:chr
| isAsciiLower char = chr shifted
| isAsciiUpper char && (shifted < ord 'A' || shifted > ord 'Z') = shift cycled char
| isAsciiUpper char = chr shifted
| otherwise = char
where shifted = ord char + amount -- e.g. 'A' (ASCII: 66) shifted 1 = 'B' (ASCII: 67), so shifted would be 67.
-- cycled: the shift amount, but cycled by 26 to start over the alphabet,
-- e.g. 'Z' (by ASCII: 90) shifted 1 is 91, so -26 to get 65, which is 'A'
-- the amount `div` amount makes sure we add or subtract 26 as needed to cycle
-- e.g. 'A' (65) shifted -1 is 64, but if we subtracted 26 from 64, it wouldn't cycle us to Z,
-- it is in the wrong direction
-- so instead we multiply by the sign of the amount: -1 - (26 * (-1 / |-1|) to get +25,
-- so 65 ('A') + 25 = 90 ('Z')
cycled = amount - ((amount `div` abs (amount)) * 26)
main = do
args <- getArgs
message <- getLine
case args of
-- `read` converts aInt from string to int
-- `map` is used to apply `shiftRight` to each char in the string `message`
["-encode", aInt] -> putStrLn $ map (shift $ read $ aInt) message
["-decode", aInt] -> putStrLn $ map (shift $ negate $ read $ aInt) message
_ -> do
progName <- getProgName
putStrLn ("Usage: " ++ progName ++ " [-encode|-decode] [0-9]")
exitFailure
我编写了以下 Haskell 程序，其目的是像 Caesar cipher:
1 import System.IO
2 import System.Environment
3 import System.Exit
4 import Data.Char
5
6 shiftRight :: Int -> Char -> Char
7 shiftRight shift char = do
8 if isAsciiLower char
9 then if (toEnum (fromEnum char + shift) :: Char) > 'z'
10 then shiftRight (shift - 26) char
11 else toEnum (fromEnum char + shift) :: Char
12 else if isAsciiUpper char
13 then if (toEnum (fromEnum char + shift) :: Char) > 'Z'
14 then shiftRight (shift - 26) char
15 else toEnum (fromEnum char + shift) :: Char
16 else char
17
18 shiftLeft :: Int -> Char -> Char
19 shiftLeft shift char = do
20 if isAsciiLower char
21 then if (toEnum (fromEnum char - shift) :: Char) < 'a'
22 then shiftLeft (shift + 26) char
23 else toEnum (fromEnum char - shift) :: Char
24 else if isAsciiUpper char
25 then if (toEnum (fromEnum char - shift) :: Char) < 'A'
26 then shiftLeft (shift + 26) char
27 else toEnum (fromEnum char - shift) :: Char
28 else char
29
30 main = do
31 args <- getArgs
32 message <- getLine
33 case args of
34 [aString, aInt] ->
35 if aString == "-encode"
36 -- `read` converts aInt from string to int
37 -- `map` is used to apply `shiftRight` to each char in the string `message`
38 then putStrLn $ show $ map (shiftRight $ read $ aInt) message
39 else
40 if aString == "-decode"
41 then putStrLn $ show $ map (shiftLeft $ read $ aInt) message
42 else do
43 putStrLn ("Second argument should be either '-decode' or '-encode'!")
44 exitFailure
45 _ -> do
46 progName <- getProgName
47 putStrLn ("Usage: " ++ progName ++ " [-encode|-decode] [0-9]")
48 exitFailure
我的ghc版本如下：
$ ghc --version
The Glorious Glasgow Haskell Compilation System, version 8.6.5
我在 macos (Catalina) 上编译 Haskell:
$ ghc Prog1d.hs -o Prog1d
Loaded package environment from $HOME/.ghc/x86_64-darwin-8.6.5/environments/default
[1 of 1] Compiling Main ( Prog1d.hs, Prog1d.o )
Linking Prog1d ...
然后我运行我的代码：
$ echo "ABCXYZabcxyz" | ./Prog1d -encode 1
"BCDYZAbcdyza"
$ echo "ABCXYZabcxyz" | ./Prog1d -encode 2
"CDEZABcdezab"
$ echo "ABCXYZabcxyz" | ./Prog1d -encode 4
"EFGBCDefgbcd"
$ echo "ABCXYZabcxyz" | ./Prog1d -encode 100
"WXYTUVwxytuv"
$ echo "ABCXYZabcxyz" | ./Prog1d -decode 1
Prog1d: Prelude.chr: bad argument: (-14)
$ echo "ABCXYZabcxyz" | ./Prog1d -decode 2
Prog1d: Prelude.chr: bad argument: (-15)
$ echo "ABCXYZabcxyz" | ./Prog1d -decode 4
Prog1d: Prelude.chr: bad argument: (-17)
$ echo "ABCXYZabcxyz" | ./Prog1d -decode 100
Prog1d: Prelude.chr: bad argument: (-35)
为什么我得到 Prelude.chr: bad argument？是什么原因造成的，我该怎么做才能解决这个问题？
我读到过其他人遇到此错误，但在他们的情况下，删除 *.hi 文件解决了问题。我删除了 Prog1d.hi（以及 Prog1d.o 和 Prog1d），但没有任何效果。我觉得这可能是由我的代码中的某些内容引起的，可能是第 41 行：
then putStrLn $ show $ map (shiftLeft $ read $ aInt) message
但是这一行就像第 38 行一样，它适用于 -encode 用例。我一定遗漏了一些明显的东西。
我是 Haskell 的新手，所以请帮助我。 我主要习惯用C++、python、Java等命令式语言编写代码，对Haskell等函数式语言的思想和语法还不熟悉
感谢阅读本文！
我发现了问题。
在我的代码中，我将一个字符转换成它对应的ASCII码。
例如fromEnum 'A'会给你65。问题是，如果您将值从 ASCII 值的 table 移开，例如-1，你导致 toEnum 的错误参数错误，我用来比较以检查偏移是否过大。
例如：
echo "A" | ./Prog1d -decode 66
这将带我到程序中的第 25 行：
25 then if (toEnum (fromEnum char - shift) :: Char) < 'A'
fromEnum char 将是 fromEnum 'A' 这将给我 65.
然后我从 65 中减去 shift 值：66。所以，65 - 66 = -1.
然后 toEnum -1 导致 Prelude.chr: bad argument: (-1).
那是因为 -1 没有 ASCII 字符，该值超出范围。
这意味着我只需比较 int 值而不是 chars 来检查我的班次是否越界。
这是更正后的代码：
import System.IO
import System.Environment
import System.Exit
import Data.Char
shiftRight :: Int -> Char -> Char
shiftRight shift char = do
if isAsciiLower char
then if (fromEnum char + shift) > 122 -- 122 is 'z' in ASCII
then shiftRight (shift - 26) char
else toEnum (fromEnum char + shift) :: Char
else if isAsciiUpper char
then if (fromEnum char + shift) > 90 -- 90 is `Z` in ASCII
then shiftRight (shift - 26) char
else toEnum (fromEnum char + shift) :: Char
else char
shiftLeft :: Int -> Char -> Char
shiftLeft shift char = do
if isAsciiLower char
then if (fromEnum char - shift) < 97 -- 97 is 'a' in ASCII
then shiftLeft (shift - 26) char
else toEnum (fromEnum char - shift) :: Char
else if isAsciiUpper char
then if (fromEnum char - shift) < 65 -- 65 is 'A' in ASCII
then shiftLeft (shift - 26) char
else toEnum (fromEnum char - shift) :: Char
else char
main = do
args <- getArgs
message <- getLine
case args of
[aString, aInt] ->
if aString == "-encode"
-- `read` converts aInt from string to int
-- `map` is used to apply `shiftRight` to each char in the string `message`
then putStrLn $ map (shiftRight $ read $ aInt) message
else
if aString == "-decode"
then putStrLn $ map (shiftLeft $ read $ aInt) message
else do
putStrLn ("Second argument should be either '-decode' or '-encode'!")
exitFailure
_ -> do
progName <- getProgName
putStrLn ("Usage: " ++ progName ++ " [-encode|-decode] [0-9]")
exitFailure
这里是正确的、期望的输出：
$ echo "ABCXYZabcxyz" | ./Prog1d -encode 1
BCDYZAbcdyza
$ echo "ABCXYZabcxyz" | ./Prog1d -encode 2
CDEZABcdezab
$ echo "ABCXYZabcxyz" | ./Prog1d -encode 4
EFGBCDefgbcd
$ echo "ABCXYZabcxyz" | ./Prog1d -encode 100
WXYTUVwxytuv
$ echo "ABCXYZabcxyz" | ./Prog1d -decode 1
ZABWXYzabwxy
$ echo "ABCXYZabcxyz" | ./Prog1d -decode 2
YZAVWXyzavwx
$ echo "ABCXYZabcxyz" | ./Prog1d -decode 4
WXYTUVwxytuv
$ echo "ABCXYZabcxyz" | ./Prog1d -decode 100
EFGBCDefgbcd
编辑 (2020-09-01)：John Purdy 给了我一些反馈，这促使我从根本上重构我的代码。这是改进后的版本：
import System.IO
import System.Environment
import System.Exit
import Data.Char
shift :: Int -> Char -> Char
shift amount char
-- `ord` is `fromEnum` but only for `Char` types; converts a Char to an Int,
-- which gives us the ASCII number for that character
-- see: https://hackage.haskell.org/package/base-4.14.0.0/docs/Data-Char.html#v:ord
| isAsciiLower char && (shifted < ord 'a' || shifted > ord 'z') = shift cycled char
-- `chr` is `toEnum` but only for `Char` types; it converts from an Int to Char
-- see: https://hackage.haskell.org/package/base-4.14.0.0/docs/Data-Char.html#v:chr
| isAsciiLower char = chr shifted
| isAsciiUpper char && (shifted < ord 'A' || shifted > ord 'Z') = shift cycled char
| isAsciiUpper char = chr shifted
| otherwise = char
where shifted = ord char + amount -- e.g. 'A' (ASCII: 66) shifted 1 = 'B' (ASCII: 67), so shifted would be 67.
-- cycled: the shift amount, but cycled by 26 to start over the alphabet,
-- e.g. 'Z' (by ASCII: 90) shifted 1 is 91, so -26 to get 65, which is 'A'
-- the amount `div` amount makes sure we add or subtract 26 as needed to cycle
-- e.g. 'A' (65) shifted -1 is 64, but if we subtracted 26 from 64, it wouldn't cycle us to Z,
-- it is in the wrong direction
-- so instead we multiply by the sign of the amount: -1 - (26 * (-1 / |-1|) to get +25,
-- so 65 ('A') + 25 = 90 ('Z')
cycled = amount - ((amount `div` abs (amount)) * 26)
main = do
args <- getArgs
message <- getLine
case args of
-- `read` converts aInt from string to int
-- `map` is used to apply `shiftRight` to each char in the string `message`
["-encode", aInt] -> putStrLn $ map (shift $ read $ aInt) message
["-decode", aInt] -> putStrLn $ map (shift $ negate $ read $ aInt) message
_ -> do
progName <- getProgName
putStrLn ("Usage: " ++ progName ++ " [-encode|-decode] [0-9]")
exitFailure