Ways to type check in javascript

Question

I am studying Javascript, and I came across a question, which I present below, I found several ways to solve the same case in Javascript, and I show you what I chose. I wonder if you know other ways, and how to better deal with this kind of situation, letting the compiler work as I did would be a really good way to deal with the issue or am I mistaken?

// Utilizando a sintaxe de módulos, AssetPackage está
// em um arquivo diferente de PackageManager

function AssetPackage(name, assets) {
    this.name = name;
    this.assets = assets;
}

function PackageManager() {
    this.packages = {};
}

I can’t let objects that aren’t packages be added a Packages, as this could make the API work badly. I imagined the following ways to solve this:

Way 1 - instanceof

// Importo AssetPackage
PackageManager.prototype.addPackage = function(assetPackage) {
    if (!this.isAssetPackage(assetPackage))
      return false;

    this.packages[assetPackage.name] = assetPackage;
}

PackageManager.prototype.addPackages = function(assetPackages) {
    if (!this.isAssetPackage(assetPackage))
      return false;

    for(let i = 0; i < assetPackages.length; i++)
        this.packages[assetPackages[i].name] = assetPackages[i];
}

PackageManager.prototype.isAssetPackage = function(val) {
    return (val instanceof AssetPackage);
}

Way 2 - Checking properties

// Não importo AssetPackage
PackageManager.prototype.addPackage = function(assetPackage) {
    if (!assetPackage.name || !assetPackage.assets)
        return false;

    this.packages[assetPackage.name] = assetPackage;
}

PackageManager.prototype.addPackages = function(assetPackages) {
    for(let i = 0; i < assetPackages.length; i++)
        if (!assetPackages.name || !assetPackages.assets)
            continue;
            this.packages[assetPackages[i].name] = assetPackages[i];
}

Way 3 - Checking properties and assigning them only

// Não importo AssetPackage
PackageManager.prototype.addPackage = function(assetPackage) {
    if (!assetPackage.name || !assetPackage.assets)
        return false;

    this.packages[assetPackage.name] = {
        name: assetPackage.name,
        assets: assetPackage.assets
    };
}

PackageManager.prototype.addPackages = function(assetPackages) {
    for(let i = 0; i < assetPackages.length; i++)
        if(!assetPackages.name || !assetPackages.assets)
            continue;
        this.packages[assetPackages[i].name] = {
            name: assetPackages[i].name,
            assets: assetPackages[i].assets
    };
}

Way 4 - Checking properties and whether an object is passed

// Não importo AssetPackage
PackageManager.prototype.addPackage = function(assetPackage) {
    if (!this.isObject(assetPackage) || !assetPackage.name || !assetPackage.assets)
        return false;

    this.packages[assetPackage.name] = {
        name: assetPackage.name,
        assets: assetPackage.assets
    };
}

PackageManager.prototype.addPackages = function(assetPackages) {
    if (!Array.isArray(assetPackages))
        return false;

    for(let i = 0; i < assetPackages.length; i++)
        if(!this.isObject(assetPackages) || !assetPackages[i].name || !assetPackages[i].assets)
            continue;
        this.packages[assetPackages[i].name] = {
            name: assetPackages[i].name,
            assets: assetPackages[i].assets
    };
}

PackageManager.prototype.isObject = function(val) {
    return (typeof val === "object" && val !== null);
}

Way 5 - constructor.name

PackageManager.prototype.addPackage = function(assetPackage) {
    if (!isAssetPackage(assetPackage))
      return false;

    this.packages[assetPackage.name] = assetPackage;
}

PackageManager.prototype.addPackages = function(assetPackages) {
    if (!isAssetPackage(assetPackage))
      return false;

    for(let i = 0; i < assetPackages.length; i++)
        this.packages[assetPackages[i].name] = assetPackages[i];
}

PackageManager.prototype.isAssetPackage = function(val) {
    return (getType(val) === 'AssetPackage');
}

PackageManager.prototype.getType = function (o) { 
    return o && o.constructor && o.constructor.name; 
}

Way 6 - The chosen one. Do not check any! let the compiler work.

// Importo AssetPackage
PackageManager.prototype.addPackage = function(assetPackage) {
    this.packages[assetPackage.name] = assetPackage.assets;
}

PackageManager.prototype.addPackages = function(assetPackages) {
    for(let i = 0; i < assetPackages.length; i++)
        this.packages[assetPackages[i].name] = assetPackages[i].assets;
}

// ou
PackageManager.prototype.addPackage = function(assetPackage) {
    this.packages[assetPackage.name] = {
      name: assetPackage.name,
      assets: assetPackage.assets
    };
}

PackageManager.prototype.addPackages = function(assetPackages) {
    for(let i = 0; i < assetPackages.length; i++)
        this.packages[assetPackages[i].name] = {
          name: assetPackages[i].name,
          assets: assetPackages[i].assets
        };
}

I chose to let the compiler work for the following reasons, if I return a "false" or whatever it is to return to the API user, it could simply ignore the false by inattention, this would cause errors in your code if it does not correctly check the return of the API functions, already letting the compiler work, your code would simply not compile and it would be warns with an error, probably of "Undefined" property or something like that, and a build error he can’t ignore. That would avoid mistakes by carelessness.