In the realm of object-oriented programming, classes can implement multiple interfaces. However, when these interfaces possess methods with identical signatures, a question arises: how does the compiler resolve which interface method is overridden?
In Java, a class implementing multiple interfaces with such overlapping methods will have only one effective implementation. The compiler does not distinguish between the methods based on their interface origin.
To understand this concept, let's consider the following example:
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In this scenario, Test implements both A and B, and both interfaces define a method named f() with the same signature. By overriding this method in Test, the class effectively overrides the f() method from both interfaces, even though only one @Override annotation is present.
This is because Java's inheritance rules dictate that methods with identical signatures are considered "override-equivalent." As a result, the compiler chooses the first overriding method it encounters, regardless of its interface origin.
Furthermore, incompatibilities can arise if the conflicting methods have different return types. In such cases, the compiler will report an error, as it's not possible to have two methods with the same signature but different return types in the same class.
To illustrate this point, let's modify the example as follows:
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Here, Gift.present() and Guest.present() have incompatible return types (void vs. boolean). As a result, Presentable cannot implement both interfaces as they violate the overriding rules.
In conclusion, when a class implements multiple interfaces with override-equivalent methods, the compiler recognizes there is only one method to implement, and the @Override annotation on the overriding method applies to all interfaces. However, incompatibilities between method signatures will lead to compilation errors.
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