Function to cache its argument's return value

Question

I want to write a function once that accepts a callback as input and returns a function. When the returned function is called the first time, it should call the callback and return that output. If it is called any additional times, instead of calling the callback again it will simply return the output value from the first time it was called.

Below is what I have tried to do. But I am not getting the results as I expected. I need to understand this concept.

function once(func) {
    let num; 
    function retFunc(x){
        num = func(x);
        return num;
    }
    return retFunc;
}

function addByTwo(input){
    return input + 2;
}

var onceFunc = once(addByTwo);

console.log(onceFunc(4));  //should log 6
console.log(onceFunc(10));  //should log 6
console.log(onceFunc(9001));  //should log 6

The num is already a good first step, but how do you remember that it was called a first time already? You always overwrite num = func(x) — Bergi
– Bergi, Commented Jan 7, 2019 at 7:29

CertainPerformance · Accepted Answer · 2019-01-07 07:29:13Z

19

You should only assign num = func(x) when num is undefined - that is, on the very first call of retFunc:

function once(func) {
  let num; 
  function retFunc(x){
    if (num === undefined) {
      num = func(x);
    }
    return num;
  }
  return retFunc;
}

function addByTwo(input){
  return input + 2;
}

var onceFunc = once(addByTwo);

console.log(onceFunc(4));  //should log 6
console.log(onceFunc(10));  //should log 6
console.log(onceFunc(9001));  //should log 6

But this isn't a guaranteed general solution - what if the passed function (addByTwo in your example) results in undefined when called? Then, the === undefined check won't work. So, it might be better to set a flag or something similar, and reassign that flag the first time the callback is called:

function once(func) {
  let num;
  let done = false;
  function retFunc(x){
    if (!done) {
      done = true;
      num = func(x);
    }
    return num;
  }
  return retFunc;
}

function returnsUndefinedOn1(input){
  return input === 1 ? undefined : input;
}

var onceFunc = once(returnsUndefinedOn1);

console.log(onceFunc(1));
console.log(onceFunc(10));
console.log(onceFunc(9001));

answered Jan 7, 2019 at 7:29

CertainPerformance

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6 Comments

CertainPerformance Over a year ago

That's the desired behavior - on an input of 1, the function returnsUndefinedOn1 returns undefined, per its name. Further calls of onceFunc will also return undefined, despite being called with an argument other than 1.

Pierre Arlaud Over a year ago

Please use typeof to compare with undefined

CertainPerformance Over a year ago

@PierreArlaud In any remotely reasonable code, it shouldn't be necessary, I think? Many global object can be redefined, that doesn't mean they shouldn't be used, that means not to reassign them.

Pierre Arlaud Over a year ago

@CertainPerformance typeof also does not trigger an exception if the variable is not declared. There is no scenario where not using typeof to compare with undefined is the better choice. My opinion on the matter is that using typeof is a good habbit to have, and "remotely reasonable codes" are no exceptions.

Ilmari Karonen Over a year ago

@PierreArlaud: In this case, getting an exception if the variable is not declared is desirable, since we're trying to test the value of a variable we know we just declared, and the exception is telling us that we made a typo in the variable name and the code is broken.

|

Mark · Accepted Answer · 2019-01-07 07:37:10Z

8

You should only call the function and assign num if it's undefined, otherwise you overwrite it every time:

function once(func) {
  let num;

  function retFunc(x) {
    num = (num === undefined) ? func(x) : num
    return num;
  }
  return retFunc;
}

function addByTwo(input) {
  return input + 2;
}

var onceFunc = once(addByTwo);

console.log(onceFunc(4)); //should log 6
console.log(onceFunc(10)); //should log 6
console.log(onceFunc(9001)); //should log 6

Note that if the function you pass in returns undefined it will be called more than once. If you need to handle that case you can set a flag to indicate whether the cached value is valid.

edited Jan 7, 2019 at 7:37

answered Jan 7, 2019 at 7:26

Mark

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2 Comments

Yang K Over a year ago

wow, you're really helpful around here aren't ya? no wonder you have 36.8k rep

Mark Over a year ago

@YangK mostly insomnia!

tswistak · Accepted Answer · 2019-01-07 07:28:03Z

3

What you're missing is removing the original function after the first execution. You should modify your code in the following way:

function once(func) {
  let num; 
  function retFunc(x){
      if (func)
          num = func(x);
      func = null;
      return num;
 }
    return retFunc;
}

function addByTwo(input){
  return input + 2;
}

var onceFunc = once(addByTwo);

console.log(onceFunc(4));  //should log 6
console.log(onceFunc(10));  //should log 6
console.log(onceFunc(9001));  //should log 6

This way you remove the function after first usage and you're just keeping the result.

answered Jan 7, 2019 at 7:28

tswistak

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1 Comment

tswistak Over a year ago

Also i'd like to say that this way is better than relying on the result, because function doesn't need to always return something. It may return undefined, but we still want to run it only once.

asleepysamurai · Accepted Answer · 2019-01-07 07:25:28Z

2

This is your problem

function retFunc(x){
        num = func(x);
        return num;
    }

You're always calling func(x). Add an if condition to check if num is undefined before calling func(x).

answered Jan 7, 2019 at 7:25

asleepysamurai

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1 Comment

Amit Kumar Over a year ago

Hi by adding if condition to check whether num is undefined my code is working now.

SchreiberLex · Accepted Answer · 2019-01-07 12:55:20Z

0

different approach: overwrite the call to func

no flags, check if result is undefined or anything required

function once(func) {
    let num;

    let internalFn = function(x) {
        num = func(x);
        internalFn = function(x) {
            return num;
        }
        return num;
    }

    function retFunc(x){
        return internalFn(x);
   }
      return retFunc;
  }

  function addByTwo(input){
    return input + 2;
  }

  var onceFunc = once(addByTwo);

  console.log(onceFunc(4));  //should log 6
  console.log(onceFunc(10));  //should log 6
  console.log(onceFunc(9001));  //should log 6

answered Jan 7, 2019 at 12:55

SchreiberLex

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