Why does the compiler give an error message when you reduce the visibility of a method while overriding it in the subclass?
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66
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.
Joey
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Joachim Sauer
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3But why are we not allowed to override a protected method and change it to private? Since the public interface is still the same, it doesn't break LSP this way. – Pacerier Aug 23 '14 at 07:32
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1The public interface does not change, but the protected does. Code in the parent class cannot access the methods of its own flesh and blood :( – Elazar Sep 07 '15 at 22:08
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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.
sepp2k
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C++ solves this problem by controlling the visibility of the inheritance relation. Also strange in Java is that super class constructors can call (virtual) subclass methods, i.e. in a scope, where the subclass instance is in an undefined state. – Sam Ginrich Jan 29 '23 at 13:02
