haskell - 这个haskell函数中的这些值来自哪里？

Question

假设我有以下功能:

sumAll :: [(Int,Int)] -> Int
sumAll xs = foldr (+) 0 (map f xs)
  where f (x,y) = x+y

sumAll [(1,1),(2,2),(3,3)]的结果将是 12 .

我不明白的是 (x,y)在哪里值来自。好吧，我知道它们来自 xs变量，但我不明白如何。我的意思是，在没有 where 关键字的情况下直接执行上面的代码，它会是这样的:

sumAll xs = foldr (+) 0 (map (\(x,y) -> x+y) xs)

我无法理解，在最上面的代码中， f 是怎么做的？变量和 (x,y)变量代表 (\(x,y) -> x+y) lambda 表达式。

Answer 1

希望这会有所帮助。关键是f应用于列表的元素，它们是成对的。

sumAll [(1,1),(2,2),(3,3)] 
      -- definition of sumAll
    = foldr (+) 0 (map f [(1,1),(2,2),(3,3)])
      -- application of map
    = foldr (+) 0 (f (1,1) : map f [(2,2),(3,3)])
      -- application of foldr
    = 0 + foldr (+) (f (1,1)) (map f [(2,2),(3,3)])
      -- application of map
    = 0 + foldr (+) (f (1,1)) (f (2,2) : map f [(3,3)])
      -- application of foldr
    = 0 + (f (1,1) + foldr (+) (f (2,2)) (map f [(3,3)]))
      -- application of f
    = 0 + (2 + foldr (+) (f (2,2)) (map f [(3,3)]))
      -- application of map
    = 0 + (2 + foldr (+) (f (2,2)) (f (3,3) : map f []))
      -- application of foldr
    = 0 + (2 + (f (2,2) + foldr (+) (f (3,3)) (map f [])))
      -- application of f
    = 0 + (2 + (4 + foldr (+) (f (3,3)) (map f [])))
      -- application of map
    = 0 + (2 + (4 + foldr (+) (f (3,3)) []))
      -- application of foldr
    = 0 + (2 + (4 + f (3,3)))
      -- application of f
    = 0 + (2 + (4 + 6))
    = 0 + (2 + 10)
    = 0 + 12
    = 12