Haskell Let 表达式求值

Question

我正在做评估 let 表达式的练习题，但我不明白这个表达式的输出。

下面是表达式:

let a = 2
    b = 1:[i * 2 | i <- b]
    f a = 1:[i * a | i <- (f a)]
in take (a+2) (f (head (tail b) ))

输出应该是[1,2,4,8]。有人可以逐步解释为什么这是输出

Answer 1

以下是分步说明:

let a = 2
    b = 1:[i * 2 | i <- b]
    f a = 1:[i * a | i <- (f a)]
in take (a+2) (f (head (tail b) ))

其中有两个不同的变量，称为 a，其中一个隐藏另一个，因此首先让我们重命名其中一个，以避免意外混淆它们:

let outer_a = 2
    b = 1:[i * 2 | i <- b]
    f a = 1:[i * a | i <- (f a)]
in take (outer_a+2) (f (head (tail b) ))

现在我们可以替换 outer_a 并计算 +:

let b = 1:[i * 2 | i <- b]
    f a = 1:[i * a | i <- (f a)]
in take 4 (f (head (tail b) ))

根据map重写列表推导式:

let b = 1:map (* 2) b
    f a = 1:map (* a) (f a)
in take 4 (f (head (tail b) ))

使用迭代而不是显式递归:

let b = iterate (* 2) 1
    f a = iterate (* a) 1
in take 4 (f (head (tail b) ))

评估b的前两个步骤:

let b = 1:2:iterate (* 2) 4
    f a = iterate (* a) 1
in take 4 (f (head (tail b) ))

替换为b:

let f a = iterate (* a) 1
in take 4 (f (head (tail (1:2:iterate (* 2) 4)) ))

评估tail:

let f a = iterate (* a) 1
in take 4 (f (head (2:iterate (* 2) 4) ))

评估head:

let f a = iterate (* a) 1
in take 4 (f 2)

替换为f a:

take 4 (iterate (* 2) 1)

评估迭代几次:

take 4 (1:2:4:8:iterate (* 2) 16)

评估采取:

[1,2,4,8]

我们就完成了。