Store the function.call method in a variable

Here is a snippet of code I'm having trouble with:

var a = function(){}
var b = a.call
b() // TypeError: b is not a function

The result of typeof b turns out to be 'function', and when logging b in the console, it shows as ƒ call() { [native code] }.

I'm puzzled by this behavior. Any insights on what might be happening here?

Answer №1

The issue with the code not functioning is due to the fact that .call() expects a function to be the reference for the this parameter when called, but you have detached it from a, breaking the connection.

As a result, because b (which is essentially Function.prototype.call) does not have a valid function specified for the this argument, an error occurs when trying to invoke it.

To fix this, you can do the following:

var b = a.call.bind(a);

This works by binding the a function as the reference for the this parameter of .call().

Another approach is using .call to call .call:

var b = a.call;
b.call(a);

This method sets the a function as the reference for the this parameter of b (which is still the Function.prototype.call method), but at the moment of invocation.

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