我只是想逐个元素地将两个列表相乘,所以我将(*)作为该函数的第一个参数传递:
apply :: Num a => (a -> a -> a) -> [a] -> [a] -> [a]
apply f xs ys = [f (xs !! i) (ys !! i) | i <- [0..(length xs - 1)]]
我可能会问一个愚蠢的问题,但我实际上搜索了很多,但却找不到.感谢你们!
haskell ×1