Function().prototype.call() implementation

Question

I ́m trying to understand how Function.prototype.call() works. I know what it does, and I can work with it, but I ́m curious about how this method is implemented.

Is it possible to write a javascript-method from scratch that does exactly the same?

• Alnitak: It's a built-in function - to the best of my knowledge there's no way to implement it in pure JS.
• Pointy: The JavaScript runtime has the ability to call functions in general, and part of that job is setting up the value of this. From that perspective there's really nothing surprising about the existence of .call() and .apply(); they pretty much have to exist anyway if the thing is going to work.
• Dominik Gabor: @Alnitak thank you, I was trying to find information about this for hours today, got kind of obsessed. You probably saved the rest of my day :D
• Alnitak: @DominikGabor with JS treating functions as first class objects then as Pointy suggests it's essential that the language have an intrinsic method (no pun intended) of invoking a function object. That method is .call (and .apply)
• tkone: FYI: github.com/mozilla/gecko-dev/blob/…

Answer 1

It's not possible to "unwrap" variable arguments without eval. If it's fine with you, you can try this:

function myCall(fun, obj) {
 var args = [].slice.call(arguments, 2);
 var arglist = args.map(function(_, n) { return "args[" + n + "]" }).join(',');
 obj._tmp = fun;
 return eval("obj._tmp(" + arglist + ")")
}

Example:

foo = {
 x: 123
}
bar = function(y) { return this.x + y }
console.log(myCall(bar, foo, 444))

• Paul Stelian: In modern JS, you can copy from the arguments array and use apply.
• Paul Stelian: So that apply is the fundamental function and call can be implemented in terms of it