I ran across this chunk of code (modified) in our application, and am confused to how it works:

    function someObject()
    {
        this.someProperty = {};
        this.foo = 
        {
            bar:
            {
                baz: function() { return "Huh?" }
            }
        };

        this.getValue = function()
        {
            return (this.someProperty && this.foo.bar && this.foo.bar.baz && this.foo.bar.baz()) || null;
        }
    }

    function test()
    {
        var o = new someObject();
        var val = o.getValue();
        alert(val);
    }

when you call the test() function, the text "Huh?" is alerted. I'm not sure how the result of getValue is returning that, I would've thought doing A && B && C && D would have returned true, rather than the value of D.

I ran across this chunk of code (modified) in our application, and am confused to how it works:

    function someObject()
    {
        this.someProperty = {};
        this.foo = 
        {
            bar:
            {
                baz: function() { return "Huh?" }
            }
        };

        this.getValue = function()
        {
            return (this.someProperty && this.foo.bar && this.foo.bar.baz && this.foo.bar.baz()) || null;
        }
    }

    function test()
    {
        var o = new someObject();
        var val = o.getValue();
        alert(val);
    }

when you call the test() function, the text "Huh?" is alerted. I'm not sure how the result of getValue is returning that, I would've thought doing A && B && C && D would have returned true, rather than the value of D.

• javascript

2 Answers 2

That happens because the Boolean Operators in JavaScript can return an operand, and not necessarily a Boolean result, e.g.:

The Logical AND operator (&&), will return the value of the second operand if the first is truthy:

true && "foo"; // "foo"

And it will return the value of the first operand if it is by itself falsy:

NaN && "anything"; // NaN
0 && "anything";   // 0

That's why in your example "Huh?" is returned, because all the preceding expressions are truthy:

alert("A" && "B" && "C" && "Huh?"); // "Huh?"
alert(true && true && true && "Huh?"); // "Huh?"

The Logical OR operator (||) has a similar behavior, it will return the value of the second operand, if the first one is falsy:

false || "bar"; // "bar"

And it will return the value of the first operand if it is by itself non-falsy:

"foo" || "anything"; // "foo"

This behavior is often used to set default values, for example:

function test (arg1) {
  arg1 = arg1 || "default value";
}

Note: Falsy values are those that coerce to false when used in a boolean context, and they are: null, undefined, NaN, 0, zero-length string, and of course false. Anything else will coerce to true.

&& and || don't neccesarily produce a boolean value.

https://developer.mozilla/en-US/docs/Web/JavaScript/Guide/Expressions_and_Operators#logical_operators