为什么在构造函数中使用super()?

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英文:

Why super() is used in Constructor?

问题

以下是您要翻译的内容:

"我在学习有关自定义 ArrayAdapter 的内容,在 GitHub 上找到了这个项目。我无法理解为什么在这里使用了 super。

public AndroidFlavorAdapter(Activity context, ArrayList<AndroidFlavor> Flavors) {
    super(context, 0, Flavors);
}

当我删除 super 时,会出现以下错误。

"X 没有适用的构造函数 '()'"

有帮助吗?"

英文:

I was learning about custom ArrayAdapter.Found this project on Github. I can't figure out why super is used here.

public AndroidFlavorAdapter(Activity context, ArrayList&lt;AndroidFlavor&gt;Flavors) {

        super(context, 0, Flavors);
  }

When I remove super this error pops up.

X There is no applicable constructor to '()’

Any help?

答案1

得分: 3

每个构造函数都需要在第一件事情做的是另一个构造函数<sup>1</sup>。有三种方法可以做到这一点:

  • 构造函数可以进行显式的super调用,可以带参数也可以不带参数。参数类型需要与在超类中声明的构造函数的签名匹配。

  • 构造函数可以进行显式的this调用。这会调用该类声明的另一个构造函数。

  • 如果没有显式的superthis调用,Java编译器会在构造函数中添加隐式的super()调用。为了使其工作,超类中需要有一个无参数的构造函数;即无参构造函数。

<sup>1 - 除了java.lang.Object,它没有超类。请注意,字节码验证器会检查这一点。如果你使用字节码汇编器创建一个不调用超类构造函数的构造函数,它将被类加载器拒绝。</sup>


所以...

> 为什么在构造函数中使用super(...)

为了显式调用超类构造函数。请注意,在这种情况下,您正在向超类构造函数传递参数。

> 当我移除super时,出现了这个错误:“没有适用于'()'的构造函数”。

这是因为编译器找不到隐式调用的超类无参构造函数,如果您没有显式的super(...)调用。

> 请问为什么super中的第二个参数是0。

超类的文档应该解释了这意味着什么。在这种情况下,第二个参数是资源ID。我不确定这是否有意义,但我曾看到有人说资源ID 0 表示null

英文:

Every constructor needs to another constructor as the first thing it does<sup>1</sup>. There are three ways to do this:

  • A constructor can make an explicit super call, either with or without parameters. The parameters types need to match the signature of a constructor declared in the superclass.

  • A constructor can make an explicit this call. This calls another constructor declared by this class.

  • If there no explicit super or this class, an implicit super() call is added to the constructor by the Java compiler. For this to work, there needs to be a constructor in the superclass with no arguments; i.e. a no-args constructor.

<sup>1 - Except for java.lang.Object which has no superclass. Note that the bytecode verifier checks this. If you use (say) a bytecode assembler to create a class with a constructor that doesn't call a superclass constructor, it will be rejected by the classloader.</sup>


So ...

> Why super(...) is used in Constructor?

To explicitly call a superclass constructor. Notice that in this case you are passing arguments to the superclass constructor.

> When I remove super this error pops up: "There is no applicable constructor to '()’"

That is because the compiler cannot find the superclasses no-args constructor that is implicitly invoked if you don't have an explicit super(...) call.

> Can you please tell me why the second parameter in super is 0.

The javadocs for the superclass should explain what that means. In this case, the 2nd parameter is a Resource ID. I'm not sure that it makes sense, but I've seen it said that Resource ID 0 means null.

答案2

得分: 0

因为基类可能正在为实例进行所需的初始化。

英文:

Because the base class is probably doing the required initialization for the instance.

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  • 本文由 发表于 2020年7月28日 10:47:56
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