What is the output for the below code?
public class Test{
public static void printValue(int i, int j, int k){
System.out.println("int");
}
public static void printValue(byte...b){
System.out.println("long");
}
public static void main(String... args){
byte b = 9;
printValue(b,b,b);
}
}
public class Test{
public static void printValue(int i, int j, int k){
System.out.println("int");
}
public static void printValue(byte...b){
System.out.println("long");
}
public static void main(String... args){
byte b = 9;
printValue(b,b,b);
}
}
A. long
B. int
C. Compilation fails
D. Compilation clean but throws RuntimeException
E. None of these
Answer: Option B
Solution(By Examveda Team)
Primitive widening uses the smallest method argument possible. (For Example if you pass short value to a method but method with short argument is not available then compiler choose method with int argument). But in this case compiler will prefer the older style before it chooses the newer style, to keep existing code more robust. var-args method is looser than widen.
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Comments ( 2 )
What is method overriding in Java?
A. Redefining a superclass method in a subclass
B. Defining a new method with the same name in a subclass
C. Making a method private in a subclass
D. Hiding methods in a superclass
What is the purpose of method overloading in Java?
A. Creating static methods
B. Hiding methods in a superclass
C. Redefining methods in a subclass
D. Defining multiple methods with the same name but different parameters
A. @OverrideMethod
B. @OverrideSuper
C. @Override
D. @OverrideParent
What happens when a subclass tries to override a final method from the superclass in Java?
A. The final method is hidden
B. It results in a compilation error
C. The final method becomes static
D. The final method is hidden
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