我不明白为什么以下功能有效:
isLongerThanN :: Integral n => n -> [a] -> Bool
isLongerThanN n xs = length xs > fromIntegral n
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但以下不是:
isLongerThanN' :: Integral n => n -> [a] -> Bool
isLongerThanN' n xs = length xs > n
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这引发了错误
Could not deduce (n ~ Int)
from the context (Integral n)
bound by the type signature for
isLongerThanN' :: Integral n => n -> [a] -> Bool
at blah.hs:140:1-35
`n' is a rigid type variable bound by
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