Question

177

How does console.log work?

rated 0 times [ 178] [ 1] / answers: 1 / hits: 23530 / 13 Years ago, thu, august 18, 2011, 12:00:00

First Example:

In the following example: http://jsfiddle.net/maniator/ScTAW/4/

I have this js:

var storage = (function () {

    var store = [];

    return {

        add: function (item) {

            store.push(item);

        },

        get: function () {

            return store;

        }

    };

}());



storage.add('hi there')

console.log(storage, storage.get(), storage.add('hi there #2'));

And here is what gets printed to the console:

Object
[hi there, hi there #2] undefined

One would think that the console should only say:

Object
[hi there] undefined

becase the second push did not happen until after the value was logged, therefore it should not be displayed.

Second Example:

In the following example: http://jsfiddle.net/maniator/ScTAW/5/

I am using the same storage variable but I log like so:

storage.add('hi there')

console.log(storage, storage.get(), (function() {

    storage.add('hi there #2');

    console.log('TESTING');

})());

What gets printed to the console is:

TESTING

Object
[hi there, hi there #2] undefined

hmmmm well that is odd now isnt it? One could expect to see:

Object
[hi there] undefined

TESTING

Why is this happening? What is going on behind the scenes of the console logging mechanism?

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reecep

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3 Years ago

answered 13 Years ago payne · Accepted Answer

In most (if not all) imperative programming languages, any arguments passed to a function call have to be evaluated before the function can be called (so called Eager evaluation). Also, they are in general evaluated in order from left to right (for C for instance it's undefined), however in both examples the order in which the arguments are evaluated does not matter. This should be pretty obvious when looking at what happens in detail:

As mentioned, before console.log can be called, storage.get() has to be executed first, returning the store array. Then storage.add('hi there #2') will be executed (or the other way round), so its result (in this case undefined, since add does not return anything) can be passed as the third argument to console.log. This means that the once console.log will be called with the arguments (storage, storage.store, undefined), the store array already contains hi there #2, hence producing the results you observe.

In the second example the reasoning is again the same, the function call is just a bit more obscured. On first look it looks there is a function passed as a 3rd argument to the console.log function; but it's actually a function call (observer the () at the end). So storage.add('hi there #2') will be executed, then console.log('TESTING') and then the undefined result from the anonymous function execution will be again passed to console.log.

If you did actually pass a function to console.log, it would print that function definition, and not execute anything. So:

storage.add('hi there')

console.log(storage, storage.get(), (function() {

    storage.add('hi there #2');

    console.log('TESTING');

}));

, without the () at the end, results in:

Object

 [hi there] function () {

    storage.add('hi there #2');

    console.log('TESTING');

}

I hope this makes things a bit clearer.