Array filter function with multiply choices using closure - Javascript

Question

I need to create a function for filter and it must have 2 choices.

1. inBetween(a, b) - which will return array between a and b
2. inArray([...]) - which will return an array of items that match with filtering array.

Something like this:

let arr = [1, 2, 3, 4, 5, 6, 7];

console.log( arr.filter(f(inBetween(3, 6))) ); // 3,4,5,6
console.log( arr.filter(f(inArray([1, 2, 10]))) ); // 1,2

I tried this function:

function f(item) {
  let result = [];

  function inBetween(from, to){
    if (item >= from && item <= to){
      result.push(item);
    }
  }

  function inArray(array){
    if (array.indexOf(item) >= 0){
      result.push(item);
    }
  }

  return result;
}

But I don't know how to attach my function into filter. It gives this error:

console.log( arr.filter(f(inBetween(3, 6))) ); // 3,4,5,6

ReferenceError: inBetween is not defined

Is it somehow possible?

Answer 1

array.filter() requires a function. If you want to pre-bind some parameters, you'll need a function that returns a function. In this case, both inBetween and inArray should return functions.

So it should be:

let arr = [1, 2, 3, 4, 5, 6, 7];

function inBetween(min, max) {
  return function(value) {
    // When this is called by array.filter(), it can use min and max.
    return min <= value && value <= max
  }
}

function inArray(array) {
  return function(value) {
    // When this is called by array.filter(), it can use array.
    return array.includes(value)
  }
}

console.log( arr.filter(inBetween(3, 6)) )
console.log( arr.filter(inArray([1, 2, 10])) )

In this case, min, max, and array close over the returned function such that when array.filter() calls the returned function, it has access to those values.

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Your inArray() functionality is already implemented by the native array.includes().