Compose函数自定义实现疑问：为何compose1报TypeError错误？

Let's break down exactly why your compose1 fails while compose works perfectly—this all comes down to how reduce is being used in each version.

First, recap your code #

Your working compose function:

const compose = (...fns) => fns.reduce((f,g) => (...args) => f(g(...args)));

Your broken compose1 function:

const compose1 = (...fns) => (...args) => fns.reduce((f,g) => f(g(...args)));

And your curried helper functions:

const multi = a => x => x*a;
const devide = a => x => x/a;
const add = a => x => x+a;

The core difference: What does reduce return at each step? #

How compose works (the correct version) #

In compose, the reduce runs immediately when you define the composite function—not when you call it later.

Each iteration of reduce returns a new function that chains the previous composite function with the next one. Let's walk through your example compose(add(6), multi(4), devide(5)):

1. First iteration: f is add(6), g is multi(4). We return (...args) => add(6)(multi(4)(...args))—a function, not a value.
2. Second iteration: f is that new function, g is devide(5). We return (...args) => [previous function](devide(5)(...args))—another function.

By the end of reduce, you have a single function that, when called with (5), will run add(6)(multi(4)(devide(5)(5)))—exactly the function pipeline you want.

How compose1 breaks (the incorrect version) #

In compose1, the reduce doesn't run until you call the final composite function (with (5)). Worse, each iteration of reduce returns the result of executing the functions, not a new function.

Let's walk through compose1(add(6), multi(4), devide(5))(5):

1. First iteration: f is add(6), g is multi(4). We calculate f(g(5)) → add(6)(multi(4)(5)) → add(6)(20) → 26. This is a number, not a function.
2. Second iteration: Now f is 26 (the result from the first step), g is devide(5). We try to run f(g(5)) → 26(devide(5)(5)) → 26(1).

You can't call a number like it's a function—hence the Uncaught TypeError: f is not a function error.

Key takeaway #

• compose uses reduce to build a reusable function pipeline—every step returns a function that can be chained further.
• compose1 uses reduce to compute a value immediately on call, but since the first step returns a non-function value, the next iteration crashes.

内容的提问来源于stack exchange，提问作者Luke