Well, the answer is hidden in the bytecode itself and it is So now the question comes if both method calls have same bytecode then how does JVM know which method to call? ) because according to compiler both methods are called on Invokevirtual #4 // Method org/programming/mitra/exercises/OverridingInternalExample$Mammal.speak:()V
Why Method overriding is called dynamic binding So in the case of method overloading compiler is able to identify the bytecode instructions and method’s address at compile time and that is why it is also known as static binding or compile time polymorphism.
#Apa itu overriding dan overloading code
Is holding, which is the object of the HumanĪnd by looking at the above code and images we can see that the bytecodes ofĪre totally different because the compiler is able to differentiate between them based on the class reference. Why Method overloading is called static bindingīut at execution time it will be called from the object which
Program’s Bytecode: executable bytecode instructions, please click one image to zoom. method references ( #Methodref), Class objects ( Constant Pool: holds almost everything which is necessary for our program’s execution e.g. Option we will get the descriptive bytecode same as our Java program.Ībove command shows the bytecode in two sections 1. Javap -verbose OverridingInternalExample.
In this section, we will try to find out physical proof of all aforementioned statements and to find them we will read the bytecode of our program which we can do by executing Once we get the doubt that how JVM is handling all this internally? or how JVM is calculating which method it should call.Īlso, we know that overloaded methods are not called polymorphic and get resolved at compile time and this is why sometimes method overloading is also known as compile time polymorphism or early/static binding.īut overridden methods get resolved at runtime time because the compiler does not know that, the object which we are assigning to our reference have overridden the method or not. Well, it is pretty simple until we keep it at the conceptual level only. But at execution time method will be called from the object which the reference is holding. Logically we can say, during compilation phase calling method is considered from the reference type. We can answer this answer in two ways, Logical way and Physical way, let's take a look at the logical way. 42: invokevirtual #9 // Method org/programming/mitra/exercises/OverridingInternalExample$Human.speak:(Ljava/lang/String )V 36: invokevirtual #7 // Method org/programming/mitra/exercises/OverridingInternalExample$Human.speak:()V 23: invokevirtual #4 // Method org/programming/mitra/exercises/OverridingInternalExample$Mammal.speak:()V 10: invokevirtual #4 // Method org/programming/mitra/exercises/OverridingInternalExample$Mammal.speak:()V Code below contains the output and and bytecode of the method callsĪnyMammal.speak() // Output - ohlllalalalalalaoaoaoa Let’s take the example of a parent class MammalĬlasses from our previous blog to understand it more clearly. In this article, we will see How Does JVM Handle Method Overloading And Overriding Internally, how JVM identifies which method should get called. In my previous article Everything About Method Overloading Vs Method Overriding, I have discussed method overloading and overriding, their rules and differences.
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