Kotlin var lazy init

Question

我想为var属性做一个懒惰的初始化.由于by lazy仅限于val属性我必须写这样的东西:

    private var currentContextProvider: ContextProvider? = null
        get() {
            if (field == null) {
                field = DefaultContextProvider()
            }
            return field
        }

现在我必须处理那些毫无意义的可控性通知:currentContextProvider?.getContext()或者说currentContextProvider!!.getContext()

难道我做错了什么？

Answer 1

您可以决定将其初始化为某个默认值，而不是将其设置为可空值，该默认值在第一次访问时将被懒惰的计算值替换：

private val noInit = "noinit"
var currentContextProvider: String = noInit
        get() = if (field == noInit) {
            synchronized(this) {
                return if (field == noInit) "lazyinit" else field
            }
        } else field

（我把它换成了ContextProvider带String）

自定义代表

以下实现了重用先前解决方案的自定义委托。可以像lazy()定义var currentContextProvider: ContextProvider by LazyMutable(){DefaultContextProvider()}

class LazyMutable<T>(val initializer: () -> T) : ReadWriteProperty<Any?, T> {
    private object UNINITIALIZED_VALUE
    private var prop: Any? = UNINITIALIZED_VALUE

    @Suppress("UNCHECKED_CAST")
    override fun getValue(thisRef: Any?, property: KProperty<*>): T {
        return if (prop == UNINITIALIZED_VALUE) {
            synchronized(this) {
               return if (prop == UNINITIALIZED_VALUE) initializer().also { prop = it } else prop as T
            }
        } else prop as T
    }

    override fun setValue(thisRef: Any?, property: KProperty<*>, value: T) {
        synchronized(this) {
            prop = value
        }
    }
}