Declaration and Access Control
What will be the output?public class Test{ public static void main(String[] args){ String value = "abc"; changeValue(value); System.out.println(value); } public static void changeValue(String a){ a = "xyz"; }}

Compilation clean but no output
xyz
None of these
Compilation fails
abc

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;public Tester(){System.out.print(this.x); // line 1Tester();}public static void Tester(){ // line 2System.out.print(this.x); // 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 invoke this)
44
Compile error at line 2 (constructors can't be static)
Compile error at line 4 (invalid argument type for method main )

ANSWER DOWNLOAD EXAMIANS APP

Declaration and Access Control
Determine output:public class InitDemo{static int i = demo();static{ System.out.print(i); }InitDemo(){System.out.print("hello1");}public static void main(String... args){ System.out.print("Hello2");} static int demo(){ System.out.print("InsideDemo");return 10;}}

Compilation error.
InsideDemo 10 Hello2
Hello2 InsideDemo 10
IllegalArgumentException is thrown at runtime.
InsideDemo 10 Hello2 hello1

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, int y){}
static void add(int...args){}
void add(Integer... args){}
static void add(int[]... args){}
static void add(int args...){}

ANSWER DOWNLOAD EXAMIANS APP