Kotlin:检查是否已初始化惰性val [英] Kotlin: Check if lazy val has been initialised

问题描述

有没有办法确定是否在Kotlin中初始化了一个懒惰的val，而没有在过程中对其进行初始化?

Is there a way to tell if a lazy val has been initialised in Kotlin without initialising it in the process?

例如，如果我有一个惰性val，则查询它是否为null会实例化它

eg if I have a lazy val, querying if it is null would instantiate it

val messageBroker: MessageBroker by lazy { MessageBroker() }
if (messageBroker == null) {
    // oops
}

我可能会使用第二个变量，但这似乎很麻烦.

I could potentially use a second variable, but that seems messy.

private var isMessageBrokerInstantiated: Boolean = false
val messageBroker: MessageBroker by lazy {
    isMessageBrokerInstantiated = true
    MessageBroker()
}

...

if (!isMessageBrokerInstantiated) {
    // use case
}

是否有一些性感的方法来确定这一点，例如if (Lazy(messageBroker).isInstantiated())?

Is there some sexy way of determining this, like if (Lazy(messageBroker).isInstantiated())?

相关(但不相同):如何检查是否有"lateinit"变量已初始化?

推荐答案

有一种方法，但是您必须访问lazy {}返回的委托对象:

There is a way, but you have to access the delegate object which is returned by lazy {}:

val messageBrokerDelegate = lazy { MessageBroker() }
val messageBroker by messageBrokerDelegate

if(messageBrokerDelegate.isInitialized())
    ...

isInitialized是接口Lazy<T>上的公共方法，这是

isInitialized is a public method on interface Lazy<T>, here are the docs.

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