javascript - 以相反顺序组成的 JS 函数

Question

从简单的函数组合开始

const fa = (x => (x + "a"));
const fb = (x => (x + "b"));
fb(fa('x'))

我尝试了一下，获得了以下代码片段，它返回“xba”而不是“xab”。

谁能解释一下为什么吗？

const fa = next => x => next(x + "a");
const fb = next => x => next(x + "b");

console.log(fb(fa(y => y))('x'));

Answer 1

让我们来分解一下:

const _fa = fa(y => y)
// _fa = x => (y => y)(x + "a")

为了避免混淆两个 x，我们将其命名为 x1

// _fa = x1 => (y => y)(x1 + "a")

现在fb将是:

// fb = next => x2 => next(x2 + "b") 

如果我们使用 fa(y => y)(即 _fa)调用 fb，我们将替换 next与_fa:

_fb = fb(fa(y => y))
// _fb = x2 => (x1 => (y => y)(x1 + "a"))(x2 + "b")

现在让我们使用参数 x2 = 'x' 来计算 _fb:

// _fb = (x1 => (y => y)(x1 + "a"))("x" + "b")
// _fb = (x1 => (y => y)(x1 + "a"))("xb")

请注意如何将 x1 => (y => y)(x1 + "a") 简化为 x1 => x1 + "a"。现在我们有:

// _fb = (x1 => x1 + "a")("xb")

现在让我们用参数 x1 = "xb" 来评估这个函数 (x1 => x1 + "a")

// _fb = "xb" + "a"
// _fb = "xba"