Declaration and Access Control Name the keyword that makes a variable belong to a class, rather than being defined for each instance of the class. abstract volatile final native static abstract volatile final native static ANSWER DOWNLOAD EXAMIANS APP
Declaration and Access Control What will be the output for the below code?public class Test{ static{ int a = 5; } public static void main(String[] args){ System.out.println(a); }} Runtime Exception 0 Compile with error 5 None of these Runtime Exception 0 Compile with error 5 None of these ANSWER DOWNLOAD EXAMIANS APP
Declaration and Access Control What is the result of compiling and running the following code?public class Tester{static int x = 4;int y = 9; public Tester(){System.out.print(this.x); // line 1printVariables();}public static void printVariables(){System.out.print(x); // line 2System.out.print(y); // line 3}public static void main(String... args) { // line 4new Tester();}} Compile error at line 1 (static x must be only accessed inside static methods) Compile error at line 3 (static methods can't make reference to non-static variables) 49 Compile error at line 4 (invalid argument type for method main) Compile error at line 2 (must access x by writing Tester.x) Compile error at line 1 (static x must be only accessed inside static methods) Compile error at line 3 (static methods can't make reference to non-static variables) 49 Compile error at line 4 (invalid argument type for method main) Compile error at line 2 (must access x by writing Tester.x) ANSWER DOWNLOAD EXAMIANS APP
Declaration and Access Control What is the result of compiling and running the following code?class Base{ private Base(){ System.out.print("Base"); }}public class test extends Base{ public test(){ System.out.print("Derived"); } public static void main(String[] args){ new test(); }} Compilation Error Exception is thrown at runtime BaseDerived Derived Compilation Error Exception is thrown at runtime BaseDerived Derived ANSWER DOWNLOAD EXAMIANS APP
Declaration and Access Control Choose all the lines which if inserted independently instead of "//insert code here" will allow the following code to compile:public class Test{ public static void main(String args[]){ add(); add(1);add(1, 2); }// insert code here} static void add(int...args){} static void add(int[]... args){} void add(Integer... args){} static void add(int args...){} static void add(int... args, int y){} static void add(int...args){} static void add(int[]... args){} void add(Integer... args){} static void add(int args...){} static void add(int... args, int y){} ANSWER DOWNLOAD EXAMIANS APP
Declaration and Access Control What will be the output after compiling and running following program code?public class Test{ static int a; public static void main(String[] args){ System.out.println("one"); call(1); } static void call(int a){ this.a=10; System.out.println("two "+a); }} None of these one two 0 one two 1 one two 10 Compile time error. None of these one two 0 one two 1 one two 10 Compile time error. ANSWER DOWNLOAD EXAMIANS APP
EXAMIANS