在Scala中将依赖对象传递给父构造函数 [英] Passing dependent objects to a parent constructor in Scala
问题描述
假设我有以下类的父类:
class A()
class B(a:A)
class C(b:B)
class BaseClass(b:B,c:C)
现在我想实现一个BaseClass的子类,它被赋予A的一个实例,并构造B和C的实例,并传递给它的超类构造函数。
如果我可以使用任意表达式,我会这样做:
b = new B(a)
c =新C(b)
super(b,c)
因为父构造函数的第二个参数取决于第一个参数的值,所以我不能看到任何方法来做到这一点,而不使用工厂函数或无意的黑客,例如:
class IntermediateSubclass(b:B)extends BaseClass(b,new C(b))
class RealSubclass是否有干净的方法?
解决方案处理这种情况的最佳方法是在您要写的BaseClass子类的随播对象中编写一个工厂方法
class A()
class B(a:A)
class C(b:B )
class BaseClass(b:B,c:C)
类SBC私有(a:A,b:B,c:C) (b,c)
对象SBC
{
def
apply(a:A):SBC = {
val b = new B )
val c = new C(b)
new SBC(a,b,c)
}
}
您可以将任何构造函数参数转换为字段而不影响任何内容(通过前缀 val
不熟悉该语法):
class A()
class B(val a:A)
类C(val b:B)
类BaseClass(val b:B,val c:C)
类SBC私有(val a:A,b: B,c:C)
extends BaseClass(b,c)
对象SBC
{
def
apply(a:A):SBC = {
val b = new B(a)
val c = new C(b)
new SBC(a,b,c)
}
}
现在可以创建 SBC
的新实例表达式的类型: SBC(aValue)
(无论是否使用 val
b
$ b
scala> val a1 = new A
a1:A = A @ 14a8f44
scala> val sbc1 = SBC(a1)
sbc1:SBC = SBC @ 7d8bb
scala> sbc1.a
res0:A = A @ 14a8f44
scala> sbc1.b
res1:B = B @ c7272
scala> sbc1.c
res2:C = C @ 178743b
Suppose I have the following class heirarchy:
class A()
class B(a:A)
class C(b:B)
class BaseClass(b:B, c:C)
Now I want to implement a subclass of BaseClass, which is given an instance of A, and constructs instances of B and C, which it passes to its superclass constructor.
If I could use arbitrary expressions, I'd do something like this:
b = new B(a)
c = new C(b)
super(b, c)
Because the second argument to the parent constructor depends on the value of the first argument, though, I can't see any way to do this, without using a factory function, or a gratuitous hack, such as :
class IntermediateSubclass(b:B) extends BaseClass(b, new C(b))
class RealSubclass(a:A) extends IntermediateSubclass(new B(a))
Is there clean way to do this?
解决方案 Probably the best way to handle this sort of situation is by writing a factory method in the companion object for the subclass of BaseClass you want to write.
class A()
class B(a:A)
class C(b:B)
class BaseClass(b:B, c:C)
class SBC private (a: A, b: B, c: C)
extends BaseClass(b, c)
object SBC
{
def
apply(a: A): SBC = {
val b = new B(a)
val c = new C(b)
new SBC(a, b, c)
}
}
You can make any of those constructor parameters into fields without affecting anything (by prefixing with val
, if you're not familiar with that syntax):
class A()
class B(val a: A)
class C(val b: B)
class BaseClass(val b: B, val c: C)
class SBC private (val a: A, b: B, c: C)
extends BaseClass(b, c)
object SBC
{
def
apply(a: A): SBC = {
val b = new B(a)
val c = new C(b)
new SBC(a, b, c)
}
}
Now new instances of SBC
can be created with this sort of expression: SBC(aValue)
(regardless of whether the val
s are used).
scala> val a1 = new A
a1: A = A@14a8f44
scala> val sbc1 = SBC(a1)
sbc1: SBC = SBC@7d8bb
scala> sbc1.a
res0: A = A@14a8f44
scala> sbc1.b
res1: B = B@c7272
scala> sbc1.c
res2: C = C@178743b
这篇关于在Scala中将依赖对象传递给父构造函数的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!