Scala setter和getter
考虑到“人”这一特征和“人”这一类:Scala setter和getter,scala,scala-repl,Scala,Scala Repl,考虑到“人”这一特征和“人”这一类: trait Person { def name: String def surname: String def married: Boolean def married_=(state: Boolean): Unit override def toString(): String = name + " " + surname + " " + married } class PersonImpl(override val name:
trait Person {
def name: String
def surname: String
def married: Boolean
def married_=(state: Boolean): Unit
override def toString(): String = name + " " + surname + " " + married
}
class PersonImpl(override val name: String,
override val surname: String,
override var married: Boolean) extends Person
我使用REPL(从命令行开始scala,只需打开一个终端并键入scala)创建trait和类。我有以下信息:
-定义特征人
-定义类PersonImpl
然后,仍然从命令行键入:
- val p:Person=新的PersonImpl1(“马里奥”,“罗西”,假)
- println(p)
- p、 已婚=真的
- 首席(p)
trait Element {
def x: Int
def y: Int
def width: Int
def height: Int
def x_:(i:Int):Unit
def y_:(i:Int):Unit
override def toString(): String = x + " " + y + " " + width + " " + height
}
class GameElement(override var x: Int,override var y: Int,override val width: Int,override val height: Int) extends Element
在命令行中键入类后,我立即遇到错误:
“变量x不覆盖任何内容”
这怎么可能呢?在Scala中,这不是推荐的方法-请查看用例类
但是为了回答你的问题,你应该把单位改成y的单位,我忘记了元素trait中的“=”,也就是:trait元素{def x:Int def y:Int def width:Int def height:Int def x:(i:Int):Unit def y:(i:Int):Unit override def toString():String=x+“”+y+“”+width+“”+height}您拼错了setter,将元素与您的工作示例进行比较,您应该会看到差异。