Haskell——不能在无穷数列中推导出(整数)

Question

powers_of :: (Integral a) => a -> [Integer]
powers_of n = sequence
where
    sequence = 1 : next sequence
    next (first : rest) = (n * first) : next rest

This get me:

Could not deduce (a ~ Integer)
from the context (Integral a)
  bound by the type signature for
             powers_of :: Integral a => a -> [Integer]
  at Triangle.hs:(10,1)-(13,61)
  `a' is a rigid type variable bound by
      the type signature for powers_of :: Integral a => a -> [Integer]
      at Triangle.hs:10:1
Expected type: [Integer]
  Actual type: [a]
In the expression: sequence
In an equation for `powers_of':
    powers_of n
      = sequence
      where
          sequence = 1 : next sequence
          next (first : rest) = (n * first) : next rest

当我去掉Integral类型类并使签名仅为Integer -> [Integer]时，一切都很好。为什么会这样?是类型推断系统的某种故障吗?

Answer 1

这绝对不是推理系统的故障。这是一个成功的推理系统。

观察到(*) :: Num a => a -> a -> a。注意到(*)的第一个论点必须与powers_of的论点统一。观察到(*)的结果必须与powers_of的结果列表的元素类型一致。

这三件事加在一起，说明powers_of的实参类型必须与结果列表的元素类型相同。

错误消息告诉您，它没有任何上下文来证明a和Integer是相同的类型，这是函数进行类型检查所需要的

Answer 2

你的类型签名应该是:

powers_of :: (Integral a) => a -> [a]

或

powers_of :: (Num a) => a -> [a]

1的类型是Num a => a, (*)的类型是Num a => a -> a -> a，所以你可以提供任何Num实例并生成该类型的列表。

没有办法隐式地将Num实例转换为Integer实例。如果需要，可以显式地这样做:

powers_of :: (Integral a) => a -> [Integer]
powers_of n = sequence
  where
    sequence = 1 : next sequence
    next (first : rest) = ((fromIntegral n) * first) : next rest