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JavaScript one-line compose (ES6)
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const compose = (...fns) => fns.reduce((f, g) => (...args) => f(g(...args))) | |
// Usage : compose functions right to left | |
// compose(minus8, add10, multiply10)(4) === 42 | |
// | |
// The resulting function can accept as many arguments as the first function does | |
// compose(add2, multiply)(4, 10) === 42 |
@teekdens sure.
We want our result function to compose its parameter functions right to left, meaning compose(f, g)(x)
should return f(g(x))
.
Let's see how compose
works.
- It casts the input functions as an array, using the ES6 array rest syntax
- It iterates over this array using
Array.prototype.reduce
. Now,Array.prototype.reduce
iterates from left to right (contrary to what we want in the end), so we have to be careful whether we returnf(g)
org(f)
on each iteration. The first parameter forreduce
is the accumulator (the value returned by the previous iteration), while the second parameter is the current value. So we want to make sure on each iteration that the current function gets called before the accumulator, thus we returnf(g)
. But since we also want the end result itself to be a function, we actually return a function which accepts any number of parameters on each iteration, so we return(...args) => f(g(...args))
.
Let's see what happens on the first example compose(minus8, add10, multiply10)
- First iteration:
f === minus8
andg === add10
, because we did not provide any initial value for the accumulator. We return the following function(...args) => minus8(add10(...args))
- Now
f === (...args) => minus8(add10(...args))
andg === multiply10
. We return(...args) => f(g(...args))
, ie.(...args) => ((...x) => minus8(add10(...x))(multiply10(...args))
. This is the final result
Let's run it on 4
.
((...args)
=> ((...x) => minus8(add10(...x))(multiply10(
...args
))
)(4)
gives us((...x) => minus8(add10(...x))(multiply10(
4
)))
- Since
multiply10(
4
)
is40
,((...x) => minus8(add10(...x))(multiply10(
4
)))
gives us((...x) => minus8(add10(...x))(
40
))
((
...x
) => minus8(add10(
...x
))(
40
))
gives usminus8(add10(
40
))
- Because
add10(
40
)
equals50
,minus8(add10(
40
))
gives usminus8(
50
)
minus8(
50
)
equals42
thanks your elaborate
Very cool piece of code - I learned a lot from it. Thank you very much for the explanation as well!
I decided to test the expanded version, works like a charm!
const multiply = (x,y) => x*y
const add6 = (x) => x+6
const minus3 = (x) => x-3
// compose(add6, minus3, multiply)(3,10)
const result = ((...arg1) => ((...arg2) => add6(minus3(...arg2)))(multiply(...arg1)))(3,10)
console.log(result) // === 33
You could probably use reduceRight
with the additional ...args
spread operator
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Nice. Thanks this helped me. However I fail to understand how this applied right to left with the usage of reduce. Can you elaborate?