Question regarding javascript call resultion on function objects

Question

I am a little bit confused regarding the way in which functions are actually called in javascript. Let's suppose we have the following sample code.

var foo = function() {
    console.log("hello world");
}
foo();

First, we create a function object which is referenced by foo, and then we invoke it. I went through the documentation which says that internally foo() actually invokes call() on the function object. This also makes sense, because, I can replace foo() with foo.call() and nothing changes for the example above.

However, let's suppose I wanted to create a function object in the following way:

class Function1 extends Function {
    call() {
        console.log("hello wold");
    }
}
const foo = new Function1();
foo.call();

When using foo.call(), the correct method is invoked. However, when I try to invoke foo(), I was hoping that the call method from Function1 would be invoked which is not the case. I guess I am thinking in the wrong direction here. I just assumed that foo.call() and foo() would be equivalent when executed on function objects which does not seem to be the case. Could somebody give me a hint on how the call resolution actually works.

It looks like you're just shadowing Function.prototype.call? — CertainPerformance
– CertainPerformance, Commented May 23, 2019 at 9:48

georg · Accepted Answer · 2019-05-23 09:49:43Z

2

When you call a function, the engine invokes [[Call]] (and not call), which is an internal method, not available in your code. If you want to intercept [[Call]], the only way is to create a Proxy with an apply hook:

function foo() {
    console.log('foo')
}

foo = new Proxy(foo, {
    apply(target) {
        console.log('foo called!')
        target()
    }
});

foo()

answered May 23, 2019 at 9:49

georg

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