Why don't logical operators (&& and ||) always return a boolean result?

Why do these logical operators return an object and not a boolean?

var _ = (obj.fn && obj.fn() ) || obj._ || ( obj._ = {} );

var _ = obj && obj._;

I want to understand why it returns result of obj.fn() (if it is defined) OR obj._ but not boolean result.


var _ = ((obj.fn && obj.fn() ) || obj._ || ( obj._ == {/* something */}))? true: false 

will return boolean.


Note that this is based on my testing. I am not to be fully relied upon.

It is an expression that does not assign true or false value. Rather it assigns the calculated value.

Let's have a look at this expression.

An example expression:

var a = 1 || 2;
// a = 1

// it's because a will take the value (which is not null) from left
var a = 0 || 2;
// so for this a=2; //its because the closest is 2 (which is not null)

var a = 0 || 2 || 1;    //here also a = 2;

Your expression:

var _ = (obj.fn && obj.fn() ) || obj._ || ( obj._ = {} );

// _ = closest of the expression which is not null
// in your case it must be (obj.fn && obj.fn())
// so you are gettig this

Another expression:

var a = 1 && 2;
// a = 2

var a = 1 && 2 && 3;
// a = 3 //for && operator it will take the fartest value
// as long as every expression is true

var a = 0 && 2 && 3;
// a = 0

Another expression:

var _ = obj && obj._;

// _ = obj._

In JavaScript, both || and && are logical short-circuit operators that return the first fully-determined “logical value” when evaluated from left to right.

In expression X || Y, X is first evaluated, and interpreted as a boolean value. If this boolean value is “true”, then it is returned. And Y is not evaluated. (Because it doesn’t matter whether Y is true or Y is false, X || Y has been fully determined.) That is the short-circuit part.

If this boolean value is “false”, then we still don’t know if X || Y is true or false until we evaluate Y, and interpret it as a boolean value as well. So then Y gets returned.

And && does the same, except it stops evaluating if the first argument is false.

The first tricky part is that when an expression is evaluated as “true”, then the expression itself is returned. Which counts as "true" in logical expressions, but you can also use it. So this is why you are seeing actual values being returned.

The second tricky part is that when an expression is evaluated as “false”, then in JS 1.0 and 1.1 the system would return a boolean value of “false”; whereas in JS 1.2 on it returns the actual value of the expression.

In JS false, 0, -0, "", null, undefined, NaN and document.all all count as false.

Here I am of course quoting logical values for discussion’s sake. Of course, the literal string "false" is not the same as the value false, and is therefore true.


In the simplest terms:

The || operator returns the first truthy value, and if none are truthy, it returns the last value (which is a falsy value).

The && operator returns the first falsy value, and if none are falsy, it return the last value (which is a truthy value).

It's really that simple. Experiment in your console to see for yourself.

"" && "Dog"    // ""
"Cat" && "Dog" // "Dog"
"" || "Dog"    // "Dog"
"Cat" || "Dog" // "Cat"


In most programming languages, the && and || operators returns boolean. In JavaScript it's different.

OR Operator:

It returns the value of the first operand that validates as true (if any), otherwise it returns the value of the last operand (even if it validates as false).

Example 1:

var a = 0 || 1 || 2 || 3;
        ^    ^    ^    ^
        f    t    t    t
             first operand that validates as true
             so, a = 1

Example 2:

var a = 0 || false || null || '';
        ^    ^        ^       ^
        f    f        f       f
                              no operand validates as true,
                              so, a = ''

AND Operator:

It returns the value of the last operand that validates as true (if all conditions validates as true), otherwise it returns the value of the first operand that validates as false.

Example 1:

var a = 1 && 2 && 3 && 4;
        ^    ^    ^    ^
        t    t    t    t
                       last operand that validates as true
                       so, a = 4

Example 2:

var a = 2 && '' && 3 && null;
        ^    ^     ^    ^
        t    f     t    f
             entire condition is false, so return first operand that validates as false,
             so, a = ''


If you want JavaScript to act the same way how other programming languages work, use Boolean() function, like this:

var a = Boolean(1 || 2 || 3);// a = true

I think you have basic JavaScript methodology question here.

Now, JavaScript is a loosely typed language. As such, the way and manner in which it treats logical operations differs from that of other standard languages like Java and C++. JavaScript uses a concept known as "type coercion" to determine the value of a logical operation and always returns the value of the first true type. For instance, take a look at the code below:

var x = mystuff || document;
// after execution of the line above, x = document

This is because mystuff is an a priori undefined entity which will always evaluate to false when tested and as such, JavaScript skips this and tests the next entity for a true value. Since the document object is known to JavaScript, it returns a true value and JavaScript returns this object.

If you wanted a boolean value returned to you, you would have to pass your logical condition statement to a function like so:

var condition1 = mystuff || document;

function returnBool(cond){
  if(typeof(cond) != 'boolean'){ //the condition type will return 'object' in this case
     return new Boolean(cond).valueOf();
  }else{ return; }
// Then we test...
var condition2 = returnBool(condition1);
window.console.log(typeof(condition2)); // outputs 'boolean' 

We can refer to the spec(11.11) of JS here of:


The production LogicalANDExpression :LogicalANDExpression &&BitwiseORExpression is evaluated as follows:

  1. Evaluate LogicalANDExpression.

2.Call GetValue(Result(1)).

3.Call ToBoolean(Result(2)).

4.If Result(3) is false, return Result(2).

5.Evaluate BitwiseORExpression.

6.Call GetValue(Result(5)).

7.Return Result(6).

see here for the spec


First, it has to be true to return, so if you are testing for truthfulness then it makes no difference

Second, it lets you do assignments along the lines of:

function bar(foo) {
    foo = foo || "default value";


var prop;
if (obj.value) {prop=obj.value;}
else prop=0;


var prop=obj.value||0;

Returning a truthy expression - rather than just true or false - usually makes your code shorter and still readable. This is very common for ||, not so much for &&.


You should think of the short-circuit operators as conditionals rater than logical operators.

x || y roughly corresponds to:

if (x) { return x } else { return y; }  

and x && y roughly corresponds to:

if (x) { return y } else { return x; }  

Given this, the result is perfectly understandable.


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