我正在努力学习Haskell,我已经使这个函数应该(我还没有实际测试过)读取变量:
import Data.Bits
import Data.Binary
getNum :: Get Int
getNum = do
l <- getWord8
let v = fromIntegral (clearBit l 7) :: Int
if testBit l 7
then do m <- getNum
return $ v .|. shiftL m 7
else return v
它编译得很好,但我希望它能够读取任何类型的整数,而不仅仅是Int
,所以我把它改成了:
import Data.Bits
import Data.Binary
getNum :: (Bits a, Integral a) => Get a
getNum = do
l <- getWord8
let v = fromIntegral (clearBit l 7) :: a
if testBit l 7
then do m <- getNum
return $ v .|. shiftL m 7
else return v
不幸的是,这会给我以下错误:
Could not deduce (Num a2) arising from a use of ‘fromIntegral’
from the context (Bits a, Integral a)
bound by the type signature for
getNum :: (Bits a, Integral a) => Get a
at test.hs:12:11-39
Possible fix:
add (Num a2) to the context of
an expression type signature: a2
or the inferred type of v :: a1
or the type signature for getNum :: (Bits a, Integral a) => Get a
In the expression: fromIntegral (clearBit l 7) :: a
In an equation for ‘v’: v = fromIntegral (clearBit l 7) :: a
In the expression:
do { l <- getWord8;
let v = ...;
if testBit l 7 then
do { m <- getNum;
.... }
else
return v }
我真的弄不清楚错误信息是想告诉我什么,我找不到任何结论性的搜索。有人可以向我解释为什么这个错误发生和如何修复它?
将:: a
从fromIntegral
行中删除:
import Data.Bits
import Data.Binary
getNum :: (Bits a, Integral a) => Get a
getNum = do
l <- getWord8
let v = fromIntegral (clearBit l 7)
if testBit l 7
then do m <- getNum
return $ v .|. shiftL m 7
else return v
解释暂时再假设下面这行:
let v = fromIntegral (clearBit l 7) :: a
此时,a
是另一个独立类型变量,与(Bits a, Integral a) => Get a
中的a
无关。因此,a
没有Num
或Bit
约束,尽管类型检查器应该正确,因为您更改了return v
。
然而,由于您缺少约束,它假设您实际上知道您在做什么,并假设一个任意类型。因为fromIntegral
需要一个Integral
实例,所以它失败了。如果您在本地再次添加这些约束,它将再次编译:
let v = fromIntegral (clearBit l 7) :: (Integral a) => a
但是,此时a
并不是函数签名中的类型变量。为此,您将需要ScopedTypeVariables
扩展。但是更好的做法是,只需放弃局部表达式签名,因为GHC将正确地推断类型。