At a quick glance, what do you think the value of 'y
' will be at each alert?
y = 1;
alert(y);
var x = function() {
alert(y);
var y = 2;
alert(y);
}
x();
When I was first beginning to work in JavaScript, my guess would've been: 'alert 1: 1', 'alert 2: 1', and 'alert 3: 2'; thinking that the global value from before the function would carry into the function until it is reassigned.
The answer is instead '1, undefined, and 2'. This result comes about because of an interesting behavior of JavaScript that I had not come across until yesterday while reading 'JavaScript Patterns'. The behavior is that the JavaScript engine takes two passes over each function scope while running the script.
In the first pass, the function is parsed and all local variables are moved to the top of the function. And the second pass runs it. So in reality, the script looks like this by the time it is run:
y = 1;
alert(y);
var x = function() {
var y;
alert(y);
y = 2;
alert(y);
}
x();
Notice that 'var y
', and not 'var y = 2
' was moved to the top of the function. By declaring an unset local variable at the top of the function, the global 'y
' variable is overwritten, causing 'alert 2' to be undefined.
In theory, as a JavaScript developer, you shouldn't be polluting the global object with variables such as 'y
' anyways, so this issue should not crop up for you. But it is good to know how the JavaScript engine actually parses your scripts. This is a good case for going ahead and just putting all of your local variables at the top of your function, since the JavaScript engine is going to do it anyways. This will reduce unexpected behaviors and is considered a best practice.
If you are interested in looking further into this type of behavior in JavaScript, checkout this blog post, which goes more in depth over JavaScript scope behavior: JavaScript-Scoping-and-Hoisting. Also pick up the book 'JavaScript Patterns' by Stoyan Stefanov. Chapter 2 alone is worth twice the book's listed price.