Top Most Java Operators and assignments Interview Questions

1. given,

class A{}

class B extends A{}

B b = null;

A a = b;

B b1= (B)a;

The above code fragment doesn’t  cause NullPointerException since we are casting amongst the members of inheritance tree. Hence b1 will have the value “null”.

Similaraly,

given,

int[] i = null;

Cloneable c = i;

i = (int [] ) c;

The above code fragment doesn’t result in runtime error(NullPointerException) since  any array extends Cloneable interface.

2. All arrays implement Cloneable interface and java.io.Serializable interface
3. given,

long [] l = {1,2,3,4,5};

int [] i  = {10,20,30};

l[0]= i[0];//legal and correct

l = i; //causes compiler error

l = (long[])i;//causes error

4. given,

final short s1 = 1;

final char c1 = 2;

short s2 = 3;

char c2 = 4;

byte b1,b2,b3,b4;

remember the following,

b1 = s1;//legal because compiler applies implicit casting since it is final variable.

b2 = c1;//legal because compiler applies implicit casting since it is final variable.

b3 = s2;//illegal because compiler doesn’t apply implicit casting.

b4 = c2;//illegal because compiler doesn’t apply implicit casting.

5. Important formula,

-x = ~x +1;

and

~x = -x-1;

6. if ,

int i1;

int i2;

int i3;

i1 = <<someNumber>>;

i2 = <<someComplexMethodName>>();

i3 = i1<<i2;

if i2 is greater than 31 then “i2&0x1f” is used in its place.

7. The assignment operators are,

=	+=	-=
*=	/=	%=
<<=	>>=	>>>=
&=	^=	|=

Note:

There is no assignment operator like &&=, ||= , ~=.

8. given,

byte b = 1;

long l = 1000;

b += l;//line 3

the above fragment will not cause compiler error since compiler applies implicit casting before executing the line number 3.the above code fragment will be executed as follows.

·         b += l;

·         b = (byte)(b+l);

·         b = (byte(1+1000);

·         b = someValue;

In general ,

E1 op= E2;

            can be written as E1 = (type)(E1 op E2);

            here type is the datatype of the operand E1 and “op” is the operator applied.

given,

interface inter{}

class B implements inter {}

class sub implements B {}

class Test

{

public static void main(String [] args)

{

            B b = new B();

            I i = b;

            sub s = (sub)b;//line 1

}}

9. At line one compiler error doesn’t occur because we applying explicit casting on members of the same inheritance tree. Thus at line 1 runtime error and not compiler error is caused. This is be cause B is superclass of sub and an instance of B cannot be converted into a type of sub class.

class Test2

{

            public static void main(String [] args)

            {

                        B[] b = {new B()};

                        sub[] s = new sub[1];

                        Object obj = b;

                        s = (sub[])obj;//line 1

            }

}

10. At line one a runtime error and not compile time error is caused.
11. given,

byte a = 1;

byte b = 2;

byte c = (byte)a+b;// Line 1

line 1 causes compiler error since casting is done only to a and not to sum of

a and b.As a result the RHS will be int and LHS will be byte which results in compiler error.