相关疑难解决方法(0)

我想测试foldl vs foldr.从我所看到的,你应该使用foldl over foldr,因为尾部递归优化.

这是有道理的.但是,运行此测试后,我很困惑:

foldr(使用时间命令时需要0.057秒):

a::a -> [a] -> [a]
a x = ([x] ++ )

main = putStrLn(show ( sum (foldr a [] [0.. 100000])))

foldl(使用time命令时需要0.089s):

b::[b] -> b -> [b]
b xs = ( ++ xs). (\y->[y])

main = putStrLn(show ( sum (foldl b [] [0.. 100000])))

很明显,这个例子很简单,但我很困惑为什么foldr击败foldl.这不应该是foldl获胜的明显案例吗？

我想要一个计算给定标记列表的平均标记的函数.这是我到目前为止所做的:

getAverageRate :: [Int]->Int
getAverageRate marks = (fromIntegral (foldr (+) 0 marks))/(fromIntegral (length marks))

我想要做的是通过使用'foldr'然后除以'length marks'来获得列表中所有标记的总和.尽管我已经使用'fromIntegral'这个错误仍然出现,我找不到问题.

No instance for (Fractional Int) arising from a use of `/'
Possible fix: add an instance declaration for (Fractional Int)
In the expression:
  (fromIntegral (foldr (+) 0 marks)) / (fromIntegral (length marks))
In an equation for `getAverageRate':
    getAverageRate marks
      = (fromIntegral (foldr (+) 0 marks))
        / (fromIntegral (length marks))