Javascript.filter () true booleans

Question

function bouncer(arr) {
  // Don't show a false ID to this bouncer.
    function a(b) {
      if(b !== false) {
        return b;
      }
    }

    arr = arr.filter(a);
    return arr;
}

bouncer([7, 'ate', '', false, 9]);

I need to return only true Boolean operators, and when I run this code, it works. However, I'm rather confused because my if statement will work, be it b! == true or b! == false. Can someone explain the reason why this works in both directions?

Answer 1

I solved a similar problem and came up with this:

function bouncer(arr) {
  return arr.filter(Boolean);
}
bouncer([7, 'ate', '', false, 9]);
// returns ['7','ate','9']

Answer 2

The easiest way to do this is:

function bouncer(arr) {
    return arr.filter(x => !!x);
}

Answer 3

, . true false :

function bouncer(arr) {
    arr = arr.filter(function(x) { console.log(x === true)
       if(x !== false) {
           return true;
       }
    });
    return arr;
}

:

function bouncer(arr) {
    return arr.filter(function(x) { console.log(x === true)
       return x !== false;
    });
}

Answer 4

.filter() ES5.

, . , !

, :

arr.filter(Boolean)

, :

arr.filter( function(x) { return Boolean(x); }); 

Boolean() , true, true, false, false!

:

var a = [1, 2, "b", 0, {}, "", NaN, 3, undefined, null, 5];

var b = a.filter(Boolean);  // [1, 2, "b", {}, 3, 5]; 

: .

Answer 5

, JavaScript "" "". (, ..)

, , :

function bouncer(arr) {
  // Don't show a false ID to this bouncer.
    function a(b) {
      if(typeof(b) === 'boolean' && !b) {
        return new Boolean(b);
      }
    }

    arr = arr.filter(a);
    return arr;
}

bouncer([7, 'ate', '', false, 9, true]);