Javascript anonymous callback function

Question

I now what's a callback fn and how it works, but I don't seem to understand how it gets called in jQuery, for example:

$('#btn').click(function(){
    alert('Test');
});

I know that when the click method had finished executing, then the callback fn gets called, but I just don't understand how is that anonymous callback fn called.

In this example:

function Greeting(function(){alert('Callback fn inside Greeting');}){
          alert('Inside Greeting fn');
        }

I'd assume that after the alert inside Greeting had been display, THEN my callback fn will get called, but it doesn't. I just see this error "uncaught syntax error: unexpected token function". My question is, how do I call the anonymous callback fn that I have as a parameter in Greeting fn? Thanks in advance.

That's because function Greeting(this section is used to pass arguments/values) into the function. — NewToJS
– NewToJS, Commented Mar 9, 2015 at 1:41

thedayturns · Accepted Answer · 2015-03-09 01:43:16Z

3

You need to do 2 things separately:

Define the function that takes an anonymous callback
Call that function, and pass in the callback.

In your second example, you're doing both at the same time - you need to separate them. Here's how you'd do it:

// Step 1: Define the function that takes an anonymous callback

function Greeting(callback) {
    console.log("inside Greeting! About to call callback.");
    callback(); // Actually call the callback function.
    console.log("Finished calling callback");
}

// Step 2: Call that function with the anonymous callback

Greeting(function() { console.log("This is the anonymous function."); });

The output of that code is:

inside Greeting! About to call callback.
This is the anonymous function.
Finished calling callback

answered Mar 9, 2015 at 1:43

thedayturns

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