Create a JavaScript function that accepts an argument and can be called both with and

Does anyone know how this code works?

function animate(t) {   
    sun.rotation.y = t/1000;
    renderer.clear();
    camera.lookAt(sun.position);
    renderer.render(scene, camera);        
    window.requestAnimationFrame(animate, renderer.domElement);
};
animate(new Date().getTime());

I noticed that the animate() function has the argument "t", but when requestAnimationFrame is called inside the function, it doesn't include "t" as an argument. Surprisingly, the program works perfectly fine without it. Can anyone explain this to me?

Answer №1

Standard scenario

In JavaScript, it is not mandatory to provide all arguments when calling a function. If an argument is not passed, the corresponding variable will default to undefined.

Moreover, you can pass more arguments than defined by the function, and then access them using the arguments object. This array-like object contains all the arguments passed to the function. For instance, in the given example, t is essentially a shortcut for arguments[0]. Thus, you can have situations like the following:

function addTwoNumbers(){
 return arguments[0] + arguments[1]
}

addTwoNumbers(2,3) //outputs 5

or like this

function retrieveTwo(a,b,c,d) {
   return 2;
}

retrieveTwo(); //outputs 2 without any issues

Specific use case

However, the animate() function is invoked within requestAnimationFrame without "t"

It's important to note that the function is not being called without any argument in this scenario. 'animate' is being passed as an argument to another function (which is expected to eventually call the function itself). When a function is referenced without () following it, it is being passed as an object rather than executed. Since functions are considered objects in JavaScript, they can be passed to functions like any other object. The following example demonstrates this concept:

function addTwo(x){
  return x+2;
}

function utilizeFunction(y,z){
   return y(z);
}

utilizeFunction(addTwo,2); //outputs 4

Answer №2

When working with JavaScript functions, you have the flexibility to call them with different numbers of arguments. These arguments are captured in an array-like object called 'arguments', while any arguments that are not passed will be marked as undefined.

Take a look at the following example function:

function calculateTotal(x, y) {
    console.log(arguments);
    x = x || 0;
    y = y || 0;
    return x + y;
}

Feel free to experiment with this function in your browser console using various arguments.

The console output will resemble the following:

function calculateTotal(x, y) {
    console.log(arguments);
    x = x || 0;
    y = y || 0;
    return x + y;
}

>calculateTotal(3, 5);
[3, 5]
8
>calculateTotal(3);
[3]
3
>calculateTotal(3, 4, 5, 6);
[3, 4, 5, 6]
7

Answer №3

It may seem like everything is "working perfectly," but in reality, there is a slight jump from the first frame to the second.

When you first call animate(), the value of t is a large integer, such as:

t = 1368990399980

This initial value is a result of passing in the return value of Date().getTime() when calling animate().

Subsequent calls to animate() are triggered as callbacks from requestAnimationFrame(), resulting in much smaller values for t:

t = 414.41499999928055
t = 431.52399999962654
t = 447.76099999944563

Each of these smaller values is due to requestAnimationFrame() providing its own parameter as the first argument for animate().

self.requestAnimationFrame = function ( callback ) {

    var currTime = Date.now();
    var timeToCall = Math.max( 0, 16 - ( currTime - lastTime ) );
    var id = self.setTimeout( function() { callback( currTime + timeToCall ); }, timeToCall );
    lastTime = currTime + timeToCall;
    return id;

};

It's worth noting that the values of t mentioned above increase in increments of approximately 16, which explains the smooth animation of your object.

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