我应该在构造函数中实例化其他类吗？

发布于 2024-10-20 06:52:09 字数 427 浏览 0 评论 0原文

最近，我看到我的一位同事在构造函数中实例化他的类，所以我开始做同样的事情，如下所示：

class FooBar{
private $model1;
private $model2;
    public function __construct() {
        $this->model1=new Model1();
            $this->model2=new Model2();
    }
}

现在我开始想，是否在需要模型的任何地方实例化模型可能会更好？

例如，函数 foo() 需要 model1，函数 bar() 需要 model2，但现在两个模型都已加载。

那么问题来了：这是实例化其他类的正确方法吗？或者我应该在函数中需要它们时实例化它们？

原文

Recently, I saw a colleague of mine instantiate his classes in a constructor, so I started doing the same, like this:

class FooBar{
private $model1;
private $model2;
    public function __construct() {
        $this->model1=new Model1();
            $this->model2=new Model2();
    }
}

And now I'm starting to wonder, if maybe instantiating the models everywhere where they are needed may be better?

E.g., function foo() needs model1 and function bar() needs model2, but now both models are loaded.

So, the question: Is this the right way to instantiate other classes? Or should I just instantiate them when I need them in a function?

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洋洋洒洒 2024-10-27 06:52:09

嗯，一如既往，没有一刀切的答案。

大多数时候，类 FooBar 聚合 $model1 和 $model2 因为它需要它们来实现其功能。在这种情况下，FooBar 不能做太多事情，除非它在这些变量中包含对象，因此在构造函数中创建它们是正确的做法。

有时不需要聚合对象来执行FooBar类的大部分功能，并且该对象的构造是一项昂贵的操作。在这种情况下，仅根据需要使用如下代码构建它是有意义的：

class FooBar {
    private $model1;
    private $model2;

    public function Frob() {
        $model = $this->getModel1();
        $model->frob();
    }

    private function getModel1() {
        if ($this->model1 === null) {
            $this->model1 = new Model1;
        }

        return $this->model1;
    }
}

但是，这只是有时。如果 FooBar 类的一半操作需要 $model1 ，另一半操作需要 $model2 ，这可能表明 FooBar code> 遇到了“让我们把所有东西都放在一个类中”的情况，应该分成两个类。

Well, as always there is no one size fits all answer.

Most of the time, class FooBar aggregates $model1 and $model2 because it needs them to fulfill its function. In this scenario there's not much that FooBar can do unless it has objects in these variables, so it's the right thing to do to create them in the constructor.

Sometimes an aggregate object is not needed to perform a large part of class FooBar's function, and the construction of that object is an expensive operation. In this case, it makes sense to only construct it on demand with code like the following:

class FooBar {
    private $model1;
    private $model2;

    public function Frob() {
        $model = $this->getModel1();
        $model->frob();
    }

    private function getModel1() {
        if ($this->model1 === null) {
            $this->model1 = new Model1;
        }

        return $this->model1;
    }
}

However, that's only sometimes. If class FooBar needs $model1 for half of its operations and $model2 for the other half, this may indicate that FooBar is suffering from a case of "let's throw everything inside one class" and should be split into two classes instead.

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