Java 如何在没有初始化和特定数量元素的情况下在 Kotlin 中创建对象数组？
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How can I create an array of objects in Kotlin without initialization and a specific number of elements?
提问by Eduardo Corona
I want to create an Array of objects with a specific number of elements in Kotlin, the problem is I don't now the current values for initialization of every object in the declaration, I tried:
我想在 Kotlin 中创建一个具有特定数量元素的对象数组，问题是我现在没有声明中每个对象的初始化当前值，我尝试过：
var miArreglo = Array<Medico>(20, {null})
in Java, I have this and is exactly what I want, but i need it in Kotlin. :
在 Java 中，我有这个，这正是我想要的，但我在 Kotlin 中需要它。：
Medico[] medicos = new Medico[20];
for(int i = 0 ; i < medicos.length; i++){
medicos[i] = new Medico();
}
What would be the Kotlink equivalent of the above Java code?
上面 Java 代码的 Kotlink 等价物是什么？
Also, I tried with:
另外，我尝试过：
var misDoctores = arrayOfNulls<medic>(20)
for(i in misDoctores ){
i = medic()
}
But I Android Studio show me the message: "Val cannot be reassigned"
但我 Android Studio 向我显示消息：“Val 无法重新分配”
采纳答案by creativecreatorormaybenot
The Kotlin equivalent of that could would be this:
Kotlin 的等价物可能是这样的：
val miArreglo = Array(20) { Medico() }
But I would strongly advice you to using Listsin Kotlin because they are way more flexible. In your case the Listwould not need to be mutable and thus I would advice something like this:
但我强烈建议您在 Kotlin 中使用Lists，因为它们更加灵活。在你的情况下，List不需要是可变的，因此我会建议这样的：
val miArreglo = List(20) { Medico() }
The two snippets above can be easily explained. The first parameter is obviouslythe Arrayor Listsize as in Java and the second is a lambda function, which is the init { ... }function. The init { ... }function can consist of some kind of operation and the last value will always be the return type and the returned value, i.e. in this case a Medicoobject.
上面的两个片段很容易解释。第一个参数显然是Java 中的ArrayorList大小，第二个参数是 lambda 函数，也就是init { ... }函数。该init { ... }函数可以由某种操作组成，最后一个值将始终是返回类型和返回值，即在这种情况下是一个Medico对象。
I also chose to use a valinstead of a varbecause List's and Array's should not be reassigned. If you want to edit your List, please use a MutableListinstead.
我还选择使用 aval而不是 a，var因为List's 和Array's 不应重新分配。如果你想编辑你的List，请改用 a MutableList。
val miArreglo = MutableList(20) { Medico() }
You can edit this list then, e.g.:
然后您可以编辑此列表，例如：
miArreglo.add(Medico())
