Laravel 容器和共享实例

Question

我想知道 Laravel 如何区分单例（共享实例）和可能在容器内被覆盖的具体实现。

容器有一个如下所示的绑定方法：

public function bind($abstract, $concrete = null, $shared = false)
{        
    // If no concrete type was given, we will simply set the concrete type to the
    // abstract type. After that, the concrete type to be registered as shared
    // without being forced to state their classes in both of the parameters.
    $this->dropStaleInstances($abstract);

    if (is_null($concrete)) {
        $concrete = $abstract;
    }

    // If the factory is not a Closure, it means it is just a class name which is
    // bound into this container to the abstract type and we will just wrap it
    // up inside its own Closure to give us more convenience when extending.
    if (!$concrete instanceof Closure) {
        $concrete = $this->getClosure($abstract, $concrete);
    }

    $this->bindings[$abstract] = compact('concrete', 'shared');

    // If the abstract type was already resolved in this container we'll fire the
    // rebound listener so that any objects which have already gotten resolved
    // can have their copy of the object updated via the listener callbacks.
    if ($this->resolved($abstract)) {
        $this->rebound($abstract);
    }
}

它还有一个调用这个函数的单例方法，但 $shared 参数总是这样：

public function singleton($abstract, $concrete = null)
{
    $this->bind($abstract, $concrete, true);
}

这里的区别在于，虽然它们都绑定在 $bindings 属性中，但单例设置它是这样的：

[concrete, true]

如果似乎没有检查它是否已经设置，这如何使它成为单例？我无法找到它是否对我们设置的 $shared 变量有任何作用。

除了这个类中还有另一个属性叫做：

/**
 * The container's shared instances.
 *
 * @var array
 */
protected $instances = [];

单身人士在这里结束似乎是合乎逻辑的，那么这究竟是什么

绑定方法示例：

https://github.com/laravel/framework/blob/5.3/src/Illuminate/Container/Container.php#L178

Answer 1

bind() 方法保存$shared here。然后make() method is using isShared()方法检查$sharedis set，然后检查它是true还是falsehere。