var __decorate = (this && this.__decorate) || function(decorators, target, key, desc) {
    var c = arguments.length,
        r = c < 3 ? target : desc === null ? desc = Object.getOwnPropertyDescriptor(target, key) : desc,
        d;
    if (typeof Reflect === "object" && typeof Reflect.decorate === "function") r = Reflect.decorate(decorators, target, key, desc);
    else
        for (var i = decorators.length - 1; i >= 0; i--)
            if (d = decorators[i]) r = (c < 3 ? d(r) : c > 3 ? d(target, key, r) : d(target, key)) || r;
    return c > 3 && r && Object.defineProperty(target, key, r), r;
};

The return value is indeed not ignored here, and is in fact used as input for a call to Object.defineProperty(). I think the documentation is inaccurate.

However, a potential pitfall I see: the decorated property ends up being defined on the prototype and is no longer an instance property because:

__decorate([
    logProperty
], Greeter.prototype, "greeting", void 0);

If you add an initializer to the greeting property (greeting = 1;) it ends up calling the setter as a result, because the compiler moves the initialization to the constructor.

Yes, I'm working on a new article which explains how it works under the hood. That's how I found this behavior.

I don't think it's a pitfall. It seems like the intended behavior.

I meant "pitfall" in the sense of being a footgun, people might not expect instance properties to get moved to the prototype just because they added a decorator 😉

It's been added to the prototype anyway:

__decorate([
  logProperty
], Greeter.prototype <=====, "greeting", void 0);

Right, that's what I said - it only goes to the prototype if you have a decorator, otherwise it's strictly an own property of the instance. Seemed surprising, even if it's intentional.

Agree. They explain why in the docs:

"This is because there is currently no mechanism to describe an instance property when defining members of a prototype, and no way to observe or modify the initializer for a property. "

Great find @NetanelBasal 👏