为什么我的箭头函数不返回数组，而是返回一个函数声明

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?

Answer 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); 

Answer 2

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

Answer 3

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); 

Answer 4

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); 

Answer 5

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); 

Answer 6

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);