Restricting Binary Output to 8 bits

Question

How Can I solve this problem in java ?

int mask = ~0;

System.out.println(Integer.toBinaryString(mask));

When I type this, the output becomes: 11111111111111111111111111111111 However I just want to get 8 bits (11111111).

Another example is,

int left = mask << 7;

System.out.println(Integer.toBinaryString(left));

Output of these two lines is: 11111111111111111111111110000000 How can I take the last 8 bit ? (10000000).

Have you tried using & 0xff to restrict the result to 8 bits? — Jon Skeet
– Jon Skeet, Commented Oct 13, 2016 at 13:08
Possible duplicate of How to convert a byte to its binary string representation — Klitos Kyriacou
– Klitos Kyriacou, Commented Oct 13, 2016 at 13:18

Pan · Accepted Answer · 2016-10-13 13:32:24Z

1

The type int has 4 bytes,every byte has 8 bits,so when you call toBinaryString of Integer class,you get 32 bit String.

You can use code below to achieve your goal.

int mask = ~0;
System.out.println(Integer.toBinaryString(mask & 0xFF));
int left = mask << 7;
System.out.println(Integer.toBinaryString(left & 0xFF));

I hope the code will help you.

answered Oct 13, 2016 at 13:32

Pan

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