Scala中奇怪的行为在类/对象初始化中

问题描述

在Scala中：

object Strange extends App {
  val x = 42
  Console.println(x) // => outputs "42", as expected
}

版本2

Version 2

object Strange extends App {
  Console.println(x) // => "0" ?!
  val x = 42
}

这也是与相同的问题， class ：

class StrangeClass {
  Console.println(x) // => still "0"
  val x = 42
}

object TestApp extends App {
  new StrangeClass()
}

正常方法的正文中没有这样的问题：

There are no such issue with regular method's body:

def nonStrangeMethod {
  Console.println(y) // => fails with "not found: value y", as expected
  y = 42
}

如果我们在val声明中添加final，行为会发生巨大变化：

And behavior changes dramatically if we'd add "final" to val declaration:

class StrangeClass {
  Console.println(x) // => "42", but at least that's expected
  final val x = 42
}

对于记录，以下Java静态（Scala的对象）对应：

For records, the following Java static (Scala's object) counterpart:

public class Strange {
    static {
        System.out.println(x);
    }
    static int x = 42;

    public static void main(String[] args) {}
}

$ b b

无法使用plain&在第3行和Java非静态（Scala的 class ）对应部分：无法引用之前定义的字段的可理解错误：

fails compilation with plain & understandable error "Cannot reference a field before it is defined" on line #3 and Java non-static (Scala's class) counterpart:

public class Strange {
    Strange() {
        System.out.println(x);
        int x = 42;
    }

    public static void main(String[] args) {
        new Strange();
    }
}

显然失败，x无法解析为变量on line＃3。

obviously fails with "x cannot be resolved to a variable" on line #3.

推荐答案

这是因为 App 其使用延迟初始化。从 中对应用程式进行编程

It's because of the App trait which uses delayed initialization. From Programming in Scala regarding the App trait:

2.9.0版本也注意到了这一点：

The 2.9.0 release notes this too:

code>不会执行Console.println（x），直到 Strange b
$ b

So Console.println(x) is not executed until Strange is ran where you get this as output:

scala> s.main(Array[String]())
0

code>之后 val x = 42 后输出：

If you add another Console.println(x) after val x = 42 then it prints out:

scala> s.main(Array[String]())
0
42

编译器知道在当前范围中存在 x ，但它会延迟其评估，直到它被执行，然后打印出 Int 这是 0 。

The compiler knows that x exists in the current scope but it delays evaluation of it until it is executed and then it prints out the default value for Int which is 0.

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