Question

i am teaching myself java and i work through the exercises in Thinking in Java.

On page 116, exercise 11, you should right-shift an integer through all its binary positions and display each position with Integer.toBinaryString.

public static void main(String[] args) {
int i = 8;
System.out.println(Integer.toBinaryString(i));
int maxIterations = Integer.toBinaryString(i).length();
int j;
for (j = 1; j < maxIterations; j++) {
    i >>= 1;
System.out.println(Integer.toBinaryString(i));
}

In the solution guide the output looks like this:

1000
1100
1110
1111

When i run this code i get this:

1000
100
10
1

What is going on here. Are the digits cut off?

I am using jdk1.6.0_20 64bit. The book uses jdk1.5 32bit.

Answer 1

It looks like there is an error in the book.

The right shift operation shifts all bits to the right, removing the least significant bit. This makes a lots more sense if you right align the results (by padding with zeros, for example).

00001000
00000100
00000010
00000001
00000000

The topmost bit shifted in is:

• 0 if your number is positive
• 1 if your number is negative.

If you want the final result to be ones then try using a negative number like -8 instead of 8.

11111111111111111111111111111000
11111111111111111111111111111100
11111111111111111111111111111110
11111111111111111111111111111111

If you use >>> instead of >> then a zero will always be shifted in, regardless of whether the number is positive or negative.