scala - 获取更新的不可变对象(immutable对象)的更短方法？

Question

我有一个带有多个参数的类，例如 class Building(val a: Int, val b: Int, val c: Int)。我必须更新的这段代码是这样的:

def updatedA(a: Int): Building = new Building(a, this.b, this.c)
def updatedB(b: Int): Building = new Building(this.a, b, this.c)

有没有更短的方法来获取像下面这样的更新对象？

def updatedA(newA: Int): Building = new { val a = newA } extends this // doesn't compile/ type is AnyRef instead of Building

Answer 1

你有没有看过copyWith() construction mechanism ？

The copyWith method takes advantage of Scala 2.8's default arguments. For each field in the class, it has a parameter which defaults to the current value. Whichever arguments aren't specified will take on their original values. We can call the method using named arguments (also in Scala 2.8).

val newJoe = joe.copyWith(phone = "555-1234")

或者查看 copy mechanism对于案例类。

/* The real power is the copy constructor that is automatically generated in the case class. I can make a copy with any or all attributes modifed by using the copy constructor and declaring which field to modify */

scala> parent.copy(right = Some(node))