帶有非實例變量助理的構造函數？

Question

我有一些看起來像這樣的類：

class Foo(val:BasicData) extends Bar(val) { 
    val helper = new Helper(val) 
    val derived1 = helper.getDerived1Value() 
    val derived2 = helper.getDerived2Value() 
} 

......只是我不想守住「助手」超越構造結束的一個實例。在Java中，我會這樣做：

public class Foo { 
    final Derived derived1, derived2; 
    public Foo(BasicData val) { 
    super(val); 
    Helper helper = new Helper(val); 
    derived1 = helper.getDerived1Value(); 
    derived2 = helper.getDerived2Value(); 
    } 
} 

那麼我該如何在Scala中做類似的事情呢？我知道用apply方法創建一個與該類同名的助手對象：我希望稍微簡潔一些。

Answer 1

最好看看javap的輸出（包括私有成員）你的結論在此之前已經向旁邊跨步任何用於中間模式匹配的Tuple2的字段。

由於斯卡拉2.8.0.RC2，這Scala代碼（躍然紙上編譯）：

class BasicData 
{ 
    def basic1: Int = 23 
    def basic2: String = "boo!" 
} 

class Helper(v: BasicData) 
{ 
    def derived1: Int = v.basic1 + 19 
    def derived2: String = v.basic2 * 2 
} 

class Bar(val v: BasicData) 

class Foo(v: BasicData) 
extends Bar(v) 
{ 
    val (derived1, derived2) = { 
    val helper = new Helper(v) 
    (helper.derived1, helper.derived2) 
    } 
} 

產生以下Foo類：

% javap -private Foo 
public class Foo extends Bar implements scala.ScalaObject{ 
    private final scala.Tuple2 x$1; 
    private final int derived1; 
    private final java.lang.String derived2; 
    public int derived1(); 
    public java.lang.String derived2(); 
    public Foo(BasicData); 
} 

Answer 2

你可以使用塊創建一個臨時的輔助VAL並返回一個元組，像這樣：

class Foo(v: BasicData) extends Bar(v) { 
    val (derived1, derived2) = { 
    val helper = new Helper(v) 
    (helper.getDerived1Value(), helper.getDerived2Value()) 
    } 
}