如果在声明变量之前使用变量，为什么不抛出ReferenceError？

Question

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?

Answer 1

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.

Answer 2

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 .

Answer 3

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');