Const sum = (...args) => args.reduce((a, b) => a +

Question

Const sum = (...args) => args.reduce((a, b) => a +

b, 0);

How I can create curry function to execute this sum function with rest params?

I would like to execute like

curry(sum)(1)()(2) => 3

0

15.06.2021

8 ответов

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decoder Автор вопроса

I have tried but sum function doesn't have length so it's returning function only

0

15.06.2021

decoder Автор вопроса

decoder
I have tried but sum function doesn't have length ...

trying but not working I am doing something wrong here

0

15.06.2021

decoder Автор вопроса

decoder

trying but not working I am doing something wrong ...

logs

0

15.06.2021

ＭＵＴＨＵ・ＫＵＭＡＲ「メーカー」 🔺

You can't, a curried function has to know its length

0

15.06.2021

decoder Автор вопроса

ＭＵＴＨＵ・ＫＵＭＡＲ「メーカー」 🔺
You can't, a curried function has to know its leng...

ohh okay I was figuring from last one hour :D

0

15.06.2021

ＭＵＴＨＵ・ＫＵＭＡＲ「メーカー」 🔺

decoder
ohh okay I was figuring from last one hour :D

Currying is taking a function like (a, b, c) -> and converting it to a -> b -> c -> So by definition it can't work if it doesn't know length

0

15.06.2021

decoder Автор вопроса

ＭＵＴＨＵ・ＫＵＭＡＲ「メーカー」 🔺
Currying is taking a function like (a, b, c) -> an...

Yeah right I though it might possible but I was stuck with unknown length of function

0

15.06.2021

ＭＵＴＨＵ・ＫＵＭＡＲ「メーカー」 🔺 · Accepted Answer

ＭＵＴＨＵ・ＫＵＭＡＲ「メーカー」 🔺

decoder

trying but not working I am doing something wrong ...

Curry will not work with variadic functions

0

15.06.2021

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Const sum = (...args) => args.reduce((a, b) => a +

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