using shift operators

[Vishnu Munnangi]

Hi all,

I am confused when using shift operators(<<,>>,>>> on hexadecimal values. Can anyone tell me how to find the easy way to deal with them.

For example consider the following code:

class EBH019 {
public static void main (String args[]) {
int i1 = 0xffffffff, i2 = i1 << 1;
int i3 = i1 >> 1, i4 = i1 >>> 1;
System.out.print(Integer.toHexString(i2) + ",");
System.out.print(Integer.toHexString(i3) + ",");
System.out.print(Integer.toHexString(i4));
}}

What is the result of attempting to compile and run the program?

a. Prints: ffffffff,ffffffff,ffffffff
b. Prints: ffffffff,ffffffff,7fffffff
c. Prints: ffffffff,7fffffff,ffffffff
d. Prints: ffffffff,7ffffffe,7ffffffe
e. Prints: fffffffe,ffffffff,ffffffff
f. Prints: fffffffe,ffffffff,7fffffff
g. Prints: fffffffe,7fffffff,ffffffff
h. Prints: fffffffe,7fffffff,7fffffff
i. Run-time error
j. Compile-time error
k. None of the above

The answer given is f.
Thanks in advance

[marc weber]

The shift operators (also known as bitshift operators) work on bits. So I think you're usually best off converting to binary to perform these operations, because this is the level on which the shift is really taking place.

Notice that each hexadecimal digit represents 4 bits (since 2 raised to the 4th power is 16). So i1=0xffffffff in binary is 1111 1111 1111 1111 1111 1111 1111 1111.

So...

i2 = i1 << 1 is 1111 1111 1111 1111 1111 1111 1111 1110, which in hexadecimal is 0xfffffffe.

i3 = i1 >> 1 is 1111 1111 1111 1111 1111 1111 1111 1111, which is still 0xffffffff (since the signed right shift fills in with whatever the most significant bit is).

i4 = i1 >>> 1 is 0111 1111 1111 1111 1111 1111 1111 1111, which is 0x7fffffff (since this is the unsigned right shift, which fills in with zeros).

[Vishnu Munnangi]

Thanks marc, now I understand very well.

Thanks for your reply.