Why doesn't my arrow function return array but instead a function declaration

Question

I tried to write a arrow function to compute the square of only the positive integers ( not include the fractions).

const realNumberArray = [4, 5.6, -9.8, 3.14, 42, 6, 8.34];
const squareList = (arr) => {
    "use strict";

    const squaredIntegers = (arr) => {
        let arrayChoosen = arr.filter(ele => ele > 0 && Number.isInteger(ele));
        return arrayChoosen.map(x => x * x);
    }

    return squaredIntegers;
};
// test your code
const squaredIntegers = squareList(realNumberArray);
console.log(squaredIntegers);

But the result is a function declaration, not a array as I expect. But when I try to modify the code in this way

const squareList = (arr) => {
  "use strict";
  let squaredIntegers = arr.filter(ele => ele > 0 && Number.isInteger(ele));
  squaredIntegers = squaredIntegers.map(val => val * val);
  return squaredIntegers;
};

Then it output the array I expect. Why in the first case it doesn't work?

In the first example the function squaredIntegers() is never called, but only the function reference is returned. So why would you expect the result to be an array in the first place? — Sirko
– Sirko, Commented Aug 2, 2018 at 6:39

Laiso · Accepted Answer · 2018-08-02 06:55:50Z

1

You need to return the returned value of your squaredIntegers function, so try with this code:

const realNumberArray = [4, 5.6, -9.8, 3.14, 42, 6, 8.34];
const squareList = (arr) => {
    "use strict";

    const squaredIntegers = (a) => {
        let arrayChoosen = a.filter(ele => ele > 0 && Number.isInteger(ele));
        return arrayChoosen.map(x => x * x);
    }

    return squaredIntegers(arr);
};
// test your code
const squaredIntegers = squareList(realNumberArray);
console.log(squaredIntegers);

edited Aug 2, 2018 at 6:55

answered Aug 2, 2018 at 6:42

Laiso

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1 Comment

fubar Over a year ago

Although you have fixed OPs code so that it works, your answer neglects to explain why their code didn't work, which was actually the original question.

Justinas · Accepted Answer · 2018-08-02 06:39:23Z

0

Because you return function name only - that is - reference to function.

Instead, you should call that function:

return squaredIntegers();

Your second example works fine because you are actually calling arr.filter() and assign it's value to squaredIntegers

answered Aug 2, 2018 at 6:39

Justinas

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1 Comment

kunquan Over a year ago

Thank you. You help me clear the concept of arrow function

Nina Scholz · Accepted Answer · 2018-08-02 06:39:36Z

0

You need to call squaredIntegers with arr for getting a new array.

return squaredIntegers(arr);

const realNumberArray = [4, 5.6, -9.8, 3.14, 42, 6, 8.34];
const squareList = (arr) => {
    "use strict";

    const squaredIntegers = (arr) => {
        let arrayChoosen = arr.filter(ele => ele > 0 && Number.isInteger(ele));
        return arrayChoosen.map(x => x * x);
    }

    return squaredIntegers(arr);
};
// test your code
const squaredIntegers = squareList(realNumberArray);
console.log(squaredIntegers);

answered Aug 2, 2018 at 6:39

Nina Scholz

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Comments

CertainPerformance · Accepted Answer · 2018-08-02 06:39:42Z

0

Because the first squareList returns the function itself, rather than the function called on the input arr. You might return its invocation instead:

const realNumberArray = [4, 5.6, -9.8, 3.14, 42, 6, 8.34];
const squareList = (arr) => {
    "use strict";

    const squaredIntegers = (arr) => {
        let arrayChoosen = arr.filter(ele => ele > 0 && Number.isInteger(ele));
        return arrayChoosen.map(x => x * x);
    }

    return squaredIntegers(arr);
};
// test your code
const squaredIntegers = squareList(realNumberArray);
console.log(squaredIntegers);

answered Aug 2, 2018 at 6:39

CertainPerformance

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Comments

Dmytro Zasiadko · Accepted Answer · 2018-08-02 06:40:20Z

0

Because you return a function when you need to return result of the function call. Check the corrected code.

const realNumberArray = [4, 5.6, -9.8, 3.14, 42, 6, 8.34];
const squareList = (arr) => {
    "use strict";

    const squaredIntegers = (arr) => {
        let arrayChoosen = arr.filter(ele => ele > 0 && Number.isInteger(ele));
        return arrayChoosen.map(x => x * x);
    }

    return squaredIntegers(arr);
};
// test your code
const squaredIntegers = squareList(realNumberArray);
console.log(squaredIntegers);

answered Aug 2, 2018 at 6:40

Dmytro Zasiadko

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Comments

Waqas Noor · Accepted Answer · 2018-08-02 06:42:27Z

0

You are returning a function rather than a function call.

Change return squaredIntegers; to return squaredIntegers(arr);

So it will become

const realNumberArray = [4, 5.6, -9.8, 3.14, 42, 6, 8.34];
const squareList = (arr) => {
    "use strict";

    const squaredIntegers = (arr) => {
        let arrayChoosen = arr.filter(ele => ele > 0 && Number.isInteger(ele));
        return arrayChoosen.map(x => x * x);
    }

    return squaredIntegers(arr);
};
// test your code
const squaredIntegers = squareList(realNumberArray);
console.log(squaredIntegers);

answered Aug 2, 2018 at 6:42

Waqas Noor

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