javascript troubleshooting issues with Array.prototype.filter when dealing with false values

//This function removes falsey values from an array
function filterFalseyValues(arr) {
  return arr.filter(ele => ele);
}
console.log(filterFalseyValues([0, "ate", "", false, 9]));

I'm confused about why just "ele" works as a condition. I am simply returning the parameter, yet it successfully filters out any falsey arguments through the function. Why does this work? Also, I am puzzled by why the line cannot be written as:
return arr.filter(ele => ele===true)
My intention is to check if an element of the array is true and return it, but this approach doesn't yield the desired result. I'm struggling to comprehend why. I can grasp the concept if I were attempting to filter like (ele => ele>5).

Answer №1

By returning only "ele", you will essentially consider everything that is not explicitly false, meaning it is not equal to 0, "", or false. This is equivalent to using ele=>ele if you utilize ele=>ele!=false

If you specify ===true, none of the elements will be true since both type and value must match true

Using ==true also does not yield any results because none of the elements meet the criteria for being true (in this case, the value 1 would qualify)

Answer №2

The effectiveness of this code lies in the use of arrow functions. By utilizing filter(ele => ele), you are essentially filtering out any falsy values in your array, keeping only truthy values.

However, if you try to apply (ele => ele>5), it will not work as expected because your array contains non-numeric values.

For a more comprehensive understanding, feel free to explore arrow function expressions on this page. Additionally, learn about falsy values in JavaScript by visiting this link.

Answer №3

Remember to add a condition for when the element is false.

function filterFalsyValues(arr) {
  return arr.filter(item => item ? true : false);
}
console.log(filterFalsyValues([0, "eat", "", false, 9]));

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