haskell - 如何纠正这个haskell对数函数？

Question

natlog x = until cond count (1,1,0)
    where 
        cond (_,val,_) = val < 0.001
        count (i,val,sum) = (i+1,(-x)^i/i,sum+val)

此函数尝试计算 log 1 + x

问题类似于What is the type signature of this Haskell function? .

错误代码:

<interactive>:1:8:
    Ambiguous type variable `t0' in the constraints:
      (Num t0) arising from the literal `1' at <interactive>:1:8
      (Integral t0) arising from a use of `natlog' at <interactive>:1:1-6
      (Fractional t0) arising from a use of `natlog'
                      at <interactive>:1:1-6
    Probable fix: add a type signature that fixes these type variable(s)
    In the first argument of `natlog', namely `1'
    In the expression: natlog 1
    In an equation for `it': it = natlog 1

Answer 1

问题是，由于 ^，您的输入需要是 Integral，由于 /，您的输入需要是 Fractional >。您可以通过对其中之一使用不同的运算符来轻松解决此问题；例如，使用 ** 而不是 ^:

natlog x = until cond count (1,1,0)
    where 
        cond (_,val,_) = val < 0.001
        count (i,val,sum) = (i+1,(-x)**i/i,sum+val)