CODEWITHSUNDEEP
javascriptvariable-declarationreferenceerrorhoisting
Drazen Bjelovuk
Drazen Bjelovuk

Reputation: 5492

Why is no ReferenceError being thrown if a variable is used before it’s declared?

I’m trying to wrap my head around the behavior of reference errors thrown in JavaScript.

In the following example, a ReferenceError is thrown at the second line, and execution breaks:

var obj = {};
obj.func1 = func2;

alert('Completed');

Whereas in this example, the code completes successfully, though obj.func1 remains undefined:

var obj = {};
obj.func1 = func2;

var func2 = function() {
    alert('func2');
};

alert('Completed');

My assumption was that an error would be thrown at the second line just the same, and when that wasn’t the case, I’d have expected obj.func1 to properly reference func2, but I’ve been double blind-sided. So what exactly is going on here?

Upvotes: 4

Views: 90

Answers (3)

Thusithz
Thusithz

Reputation: 884

Your func2 variable is not visible. That's why obj.func1 remains undefined.

var obj = {};
var func2 = function() {
    alert('func2');
    return "Test";
};
    
obj.func1 = func2;
   
alert('Completed');

Upvotes: 1

Barmar
Barmar

Reputation: 782130

This is due to Javascript variable declaration "hoisting".

A variable declared with var is visible everywhere in the function, so there's no Reference Error. However, it doesn't actually receive its value until you execute the statement that initializes it. So your second example is equivalent to:

var func2;
var obj = {};
obj.func1 = func2;

func2 = function() {
    alert('func2');
};

alert('Completed');

In this rewrite, you can see that the variable exists when you perform the assignment to obj.func1. But since it doesn't yet have a value, you assign undefined to obj.func1. Assigning to func2 later doesn't change that.

Upvotes: 4

Amadan
Amadan

Reputation: 198446

var is hoisted; the variable exists throughout the current scope. Thus, the second example is equivalent to:

var obj;
var func2;

obj = {};
obj.func1 = func2;
func2 = function() {
    alert('func2');
}

alert('Completed');

Thus, when you do the assignment, The name func2 is known, but undefined. In the first example, it is unknown, which raises ReferenceError.

Upvotes: 3

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