Function for adding numbers in Javascript with currying

Question

Function for adding numbers in Javascript with currying

Can you help me solve this interesting problem? I need a javascript function that can return the sum of all arguments passed to it, even when called multiple times.

I have identified different ways in which the function can be called -

sum(1, 2, 3, 4);
sum(1, 2)(3, 4);
sum(1, 2)(3)(4);
sum(1, 2, 3)(4);
sum(1)(2, 3, 4);

All these variations should work correctly and return the total of 10.

I have made an attempt at solving this issue, but my code only works for the first two function calls sum(1, 2, 3, 4) and sum(1, 2)(3, 4). It fails for the other cases.

const arr = [];

function sum(...args) {
  if (args.length === 4) {
    return args.reduce((acc, curr) => {
      return (acc = acc + curr);
    }, 0);
  } else {
    arr.push(...args);
    return function(...args) {
      arr.push(...args);
      return sum(...arr);
    };
  }
}

I would appreciate any assistance with this problem as it's really getting frustrating.

Thank you!

javascript recursion closures

Answer 1

Answer №1

Great job, you're almost there. A clever way to approach this is by utilizing the .bind method to create a function that captures the initial arguments if less than four are provided.

Here's an example of how you can implement this:

function calculateTotal(...args) {
  if (args.length >= 4) {
    return args.reduce((acc, curr) => {
      return acc + curr;
    }, 0);
  } else {
    // Bind the available arguments and return the function:
    return calculateTotal.bind(null, ...args)
  }
}

console.log(calculateTotal(1, 2, 3, 4));
console.log(calculateTotal(1, 2)(3, 4));
console.log(calculateTotal(1, 2)(3)(4));
console.log(calculateTotal(1, 2, 3)(4));
console.log(calculateTotal(1)(2, 3, 4));

Additionally, consider updating the check condition to ensure it doesn't lead to infinite currying by changing to >= 4.

Answer 2

Great job, you're almost there. A clever way to approach this is by utilizing the .bind method to create a function that captures the initial arguments if less than four are provided.

Here's an example of how you can implement this:

function calculateTotal(...args) {
  if (args.length >= 4) {
    return args.reduce((acc, curr) => {
      return acc + curr;
    }, 0);
  } else {
    // Bind the available arguments and return the function:
    return calculateTotal.bind(null, ...args)
  }
}

console.log(calculateTotal(1, 2, 3, 4));
console.log(calculateTotal(1, 2)(3, 4));
console.log(calculateTotal(1, 2)(3)(4));
console.log(calculateTotal(1, 2, 3)(4));
console.log(calculateTotal(1)(2, 3, 4));

Additionally, consider updating the check condition to ensure it doesn't lead to infinite currying by changing to >= 4.

Answer 3

Answer №2

When it comes to currying, there is a specific behavior that doesn't always play nicely with variadic functions due to the uncertainty of arity. However, if you specify an arity in your problem (e.g., 4), it becomes possible to determine when to return a result.

const curryN = (n, f, ...xs) =>
  (...ys) =>
    ys.length >= n
      ? f(...xs, ...ys)
      : curryN(n - ys.length, f, ...xs, ...ys)
      
const add = (...numbers) =>
  numbers.reduce((a, b) => a + b, 0)
  
const curryAdd =
  curryN(4, add)
  
console.log(
  curryAdd(1)(2)(3)(4), // 10
  curryAdd(1, 2)(3, 4), // 10
  curryAdd(1, 2, 3)(4), // 10
  curryAdd(1)(2, 3, 4), // 10
  curryAdd(1, 2, 3, 4)  // 10
)

Although this approach can be fragile for designing programs and deviates from true currying that only accepts one argument per application, partial application offers a much more reliable behavior.

const partial = (f, ...xs) =>
  (...ys) =>
    f(...xs, ...ys)
    
const add = (...numbers) =>
  numbers.reduce((a, b) => a + b, 0)
  
console.log(
  partial(add, 1)(2, 3, 4), // 10
  partial(add, 1, 2)(3, 4), // 10
  partial(add, 1, 2, 3)(4)  // 10
)

Please refer to this related answer for further insights.

Answer 4

When it comes to currying, there is a specific behavior that doesn't always play nicely with variadic functions due to the uncertainty of arity. However, if you specify an arity in your problem (e.g., 4), it becomes possible to determine when to return a result.

const curryN = (n, f, ...xs) =>
  (...ys) =>
    ys.length >= n
      ? f(...xs, ...ys)
      : curryN(n - ys.length, f, ...xs, ...ys)
      
const add = (...numbers) =>
  numbers.reduce((a, b) => a + b, 0)
  
const curryAdd =
  curryN(4, add)
  
console.log(
  curryAdd(1)(2)(3)(4), // 10
  curryAdd(1, 2)(3, 4), // 10
  curryAdd(1, 2, 3)(4), // 10
  curryAdd(1)(2, 3, 4), // 10
  curryAdd(1, 2, 3, 4)  // 10
)

Although this approach can be fragile for designing programs and deviates from true currying that only accepts one argument per application, partial application offers a much more reliable behavior.

const partial = (f, ...xs) =>
  (...ys) =>
    f(...xs, ...ys)
    
const add = (...numbers) =>
  numbers.reduce((a, b) => a + b, 0)
  
console.log(
  partial(add, 1)(2, 3, 4), // 10
  partial(add, 1, 2)(3, 4), // 10
  partial(add, 1, 2, 3)(4)  // 10
)

Please refer to this related answer for further insights.

Function for adding numbers in Javascript with currying

Answer №1

Answer №2

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