Why does Array.filter(Number) filter zero out in JavaScript?
Because 0 is one of the many falsy values in javascript
All these conditions will be sent to else blocks:
if (false)if (null)if (undefined)if (0)if (NaN)if ('')if ("")if (``)From the Array.prototype.filter() documentation:
filter()calls a providedcallbackfunction once for each element in an array, and constructs a new array of all the values for which callback returns a value that coerces to true
In your case the callback function is the Number. So your code is equivalent to:
[-1, 0, 1, 2, 3, 4, Number(0), '', 'test'].filter(a => Number(a)) // Number(0) -> 0// Number(Number(0)) -> 0// Number('') -> 0// Number('test') -> NaNWhen filter function picks truthy values (or values that coerces to true), the items which return 0 and NaN are ignored. So, it returns [-1, 1, 2, 3, 4]
To prevent a falsy zero from filtering, you could use another callback for getting only numerical values: Number.isFinite
console.log([-1, 0, 1, 2, 3, 4, Number(0), '', 'test'].filter(Number.isFinite))
Expected behavior
This behavior isn't unique to using Number as the filter function. A filter function that simple returns the 0 value would also remove it from the list.
var a = [-1, 0, 1, 2, 3, 4, Number(0), '', 'test'].filter(v => v)console.log(a); // [-1, 1, 2, 3, 4, "test"]This is because Number isn't specifically a filter function, it's primarily a type-casting function (and a class constructor, but not a very useful one). So when a number (like 0) is passed to Number, it just returns that number.
Array.prototype.filter removes values that are falsy. In JavaScript, the following are falsy and thus removed by filter.
falsenullundefined0NaN''""``(For complicated backwards compatibility reasons MDN goes into, document.all is also falsy in many browsers despite being an object, but that's a side-note)