How to uncurry a function up to depth n in JavaScript ?
Last Updated :
10 Jan, 2023
The following approach covers how to uncurry a function up to depth n in JavaScript. Uncurry a function is a process of wrapping function arguments and passing all the arguments at once. Uncurry up-to-depth n means we have passed only n arguments in all arguments.
This can be done by following methods:
Using reduce() Method: Reduce method is used to apply an operation on all the elements of the list and returns the result. It takes a callback that applies to all the elements. To slice args up to n we use the slice() method. The slice() method extracts part of the elements from the list. The slice() takes starting index and end index up to what we want to extract. Starting index is by default 0 if not provide any args. End index we can provide up to the end of the list.
Example:
Javascript
<script>
let uncurry = (f, n) => (...args) => {
let func = f => a => a.reduce((l, m) => l(m), f)
if (n > args.length)
return " less args provide " ;
return func(f)(args.slice(0, n));
}
let n = 3;
const sum = x => y => z => x + y + z;
let t = uncurry(sum, n);
let ans = t(1, 2, 4, 5);
console.log(`Sum of ${n} args are ${ans}`);
</script>
|
Output:
Sum of 3 args are 7
Using for-of loop Method: In this method, we use for-of loop. The for-of loop is used to iterate over all the elements of the list. Following are the implementation of the method:
- Define n and sum function that add all provided arguments.
- Call uncurry function with the sum, n, and all list of numbers.
- The uncurry function checks if n is greater than arguments then returns fewer args provide.
- Else it iterates the overall argument and applies function sum to each argument.
- In for loop, we count iteration. When iteration becomes equal to n break the loop.
- At the end returns f which sum of all passed elements.
Example:
Javascript
const sum = x => y => z => x + y + z;
const uncurry = f => n => {
let k = 0;
return (...args) => {
if (n > args.length) {
return "n" ;
}
for (let arg of args) {
if (k === n)
break ;
k++
f = f(arg);
}
return f
};
}
let n = 3;
const uncurried = uncurry(sum)(n);
let ans = uncurried(0, 1, 7, 2, 4);
console.log(`Sum of ${n} args is ${ans}`);
|
Output:
Sum of 3 args is 8
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