scala覆盖受保护的成员函数

trait A {
    protected def g( i: Int ) = println( i )
    def f( i: Int ) = println( i )
}

我创建了一个对象a1

val a1= new A {
    override def f( i: Int ) = super.f(i)
    override def g( i: Int ) = super.g(i)
    def h( i: Int ) = super.g(i)
}

并试图调用方法

a1.f(1)
a1.g(3)    // this call fails
a1.h(5)

对于a1.g(3)我收到此错误：

<console>:10: error: method g in trait A cannot be accessed in A{def h(i: Int): Unit}
   Access to protected method g not permitted because
   enclosing object $iw is not a subclass of 
   trait A where target is defined
                      a1.g(3)    // this call fails

但是当我定义一个特性A2扩展A并覆盖方法f和g时,创建它的一个实例并调用方法,一切正常

trait A2 extends A {
    override def f( i: Int ) = super.f(i)
    override def g( i: Int ) = super.g(i)
    def h( i: Int ) = super.g(i)
}
val a2= new A2 {}

a2.f(2)
a2.g(4)
a2.h(6)

为什么两者之间存在差异

val a1= new A {
    override def g( i: Int ) = super.g(i)
}

和

trait A2 extends A {
    override def g( i: Int ) = super.g(i)
}
val a2= new A2 {}

？

谢谢！