将MutableLiveData公开为LiveData的正确方法? [英] Correct way to expose MutableLiveData as LiveData?

问题描述

请考虑以下公开MutableLiveData的方法:

Consider the following ways to expose MutableLiveData:

方法A

class ThisViewModel : ViewModel() {

    private val _someData = MutableLiveData(true)
    val someData: LiveData<Boolean>
        get() = _someData
}

// Decompiled Kotlin bytecode

public final class ThisViewModelDecompiled extends ViewModel {

   private final MutableLiveData _someData = new MutableLiveData(true);

   @NotNull
   public final LiveData getSomeData() {
      return (LiveData)this._someData;
   }
}

方法B

class ThatViewModel : ViewModel() {

    private val _someData = MutableLiveData(true)
    val someData: LiveData<Boolean> = _someData
}

// Decompiled Kotlin bytecode

public final class ThatViewModelDecompiled extends ViewModel {

   private final MutableLiveData _someData = new MutableLiveData(true);

   @NotNull
   private final LiveData someData;

   @NotNull
   public final LiveData getSomeData() {
      return this.someData;
   }

   public ThatViewModel() {
      this.someData = (LiveData)this._someData;
   }
}

是否有理由使用方法B而不是方法A?

Is there a reason to use Method B over Method A?

推荐答案

从 Java 角度来看，方法A 在类中的字段较少，因此""高效的.从 Kotlin 的角度来看，方法B 表示得更清楚一点，即非可变属性是对可变变量的直接引用.同样， Kotlin 足够聪明，可以本地访问该字段，而不是使用getter方法.

From the Java perspective, Method A has one less field in the class, thus is "more" efficient. From the Kotlin perspective, Method B denotes a bit more clearly, that the non-mutable property is a direct reference to the mutable one. Also Kotlin is clever enough to locally access the field rather than the getter method.

是否有理由使用方法B而不是方法A?

Is there a reason to use Method B over Method A?

通常只是口味问题.从微观优化的角度来看，它取决于您是否还要在类本身中使用此引用.

In general is merely a matter of taste. Looking at it from a micro-optimization perspective it depends on whether or not you'd also use this reference within the class itself.

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