Question

I am looking at the example here Using apply to chain constructors

I understand it except for this line:

fNewConstr.prototype = fConstructor.prototype;

Why is it necessary and why does it not make it lose the function that was just defined for fNewConstr?

Function.prototype.construct = function (aArgs) {
    var fConstructor = this, fNewConstr = function () { fConstructor.apply(this, aArgs); };
    // Why doesn't fNewConstr.prototype get completely overwritten?
    fNewConstr.prototype = fConstructor.prototype;
    return new fNewConstr();
};



function MyConstructor () {
    for (var nProp = 0; nProp < arguments.length; nProp++) {
        this["property" + nProp] = arguments[nProp];
    }
}

var myArray = [4, "Hello world!", false];
var myInstance = MyConstructor.construct(myArray);

alert(myInstance.property1); // alerts "Hello world!"
alert(myInstance instanceof MyConstructor); // alerts "true"
alert(myInstance.constructor); // alerts "MyConstructor"

Answer 1

If you mean, why doesn’t fNewConstr (the function) get overwritten when you write

fNewConstr.prototype = ...;

…the answer is because nothing is overwriting it. That code just sets the prototype property of the function.

If your question is: Why doesn’t fNewConstr get recreated each time construct is called, the answer is: It is.