Applying recursion for deep-nested higher-order functions

Question

I am solving Kolodny's Javascript exercises (here), specifically the 'value' exercise.

The problem requires me to create a function, fn where fn(value) will return an answer. If value is a scalar (i.e. 4), it will return 4.

If value is a function, it will return the return value of that function. If the value is a nested function, it will return the value of the deep-nested function. For example:

var fn = function() {
  return function() {
    return 4;
   };
};
assert.equal(value(fn), 4);

I have solved the problem naively using the following:

exports.value = (val) => {
  if (typeof val == 'function') {
    if (typeof val() == 'function') {
      if (typeof val()() =='function') {
        if (typeof val()()() =='function') {
          return false
        } else {
          return val()()();
        }
      } else {
        return val()();
      }
    } else {
      return val();
    }
  } else {
    return val;
  }
}

This code begs for reusability. Is there a way to use recursion to call n-number of deep-nested, anonymous functions?

Bergi · Accepted Answer · 2018-05-24 16:22:50Z

5

const value = v => typeof v === "function" ? value(v()) : v;

Just call value again with the result of the function.

A bit more readable:

function value(v) {
  if (typeof v === "function") {
    return value(v()); // !!!
  } else {
    return v;
  }
}

edited May 24, 2018 at 16:22

Bergi

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answered May 24, 2018 at 16:16

Jonas Wilms

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4 Comments

Alejandro Over a year ago

When testing it, I get the error "value is not defined." I'm assuming because value exports an anonymous function? like so exports.value = (v) => {...}

Jonas Wilms Over a year ago

@agomez yes, then do const value = exports.value = v => ...

Alejandro Over a year ago

that works great. I hadn't thought of exporting a value like that before, thanks. So that means "value" is set to an exports object, which returns an anonymous function, right? If so, wouldn't the notation to call it be value.exports()? Just wondering

Jonas Wilms Over a year ago

@agomez no, it would be exports.value() otherwise :)

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