问题
Why does the compiler give an error message when you reduce the visibility of a method while overriding it in the subclass?
回答1:
Because every instance of the subclass still needs to be a valid instance of the base class (see Liskov substitution principle).
If the subclass suddenly has lost one property of the base class (namely a public method for example) then it would no longer be a valid substitute for the base class.
回答2:
Because if this was allowed, the following situation would be possible:
Class Sub inherits from class Parent. Parent has a public method foo, Sub makes that method private. Now the following code would compile fine, because the declared type of bar is Parent:
Parent bar = new Sub();
bar.foo();
However it is not clear how this should behave. One possibility would be to let it cause a runtime error. Another would be to simply allow it, which would make it possible to call a private method from outside, by just casting to the parent class. Neither of those alternatives are acceptable, so it is not allowed.
回答3:
Because subtypes have to be usable as instances of their supertype.
来源：https://stackoverflow.com/questions/1600667/why-cant-you-reduce-the-visibility-of-a-method-in-a-java-subclass