I have an array of bytes. I want to simulate corrupt data by changing one bit, two, and three bits.
How do I do this?
-
The "Related" column on the right shows the following possible duplicates: stackoverflow.com/questions/1956160/…, stackoverflow.com/questions/560956/…Péter Török– Péter Török2010-04-24 15:31:29 +00:00Commented Apr 24, 2010 at 15:31
Add a comment
|
3 Answers
Use the xor (^) operator:
// Generate some data.
byte[] b = new byte[1024];
new Random().nextBytes(b);
// Simulate corruption of data by flipping the low 3 bits.
for (int i = 0; i < b.length; i++) {
//b[i] ^= 0x1; // change one bit
//b[i] ^= 0x3; // change two bits
b[i] ^= 0x7; // change three bits
}
Comments
The usual way to set a bit in most languages is to 'or'(|) it with a bit mask with only that bit set, and the usual way to unset it is to 'and'(&) it with a bit mask without that bit set.
So make 8 set bit masks
byte setbm1 = 1; //00000001
byte setbm2 = 1 >>1;
byte setbm3 = 1>>2;
...
...
byte setbm8 = 1>>7; //10000000
and 8 'and' bit masks
byte unsetbm1 ^= setbm1; //11111110
..
..
byte unsetbm1 ^= setbm8; //01111111
To set the first bit use (assume array is a byte array and i is an integer)
array[i] |= setbm1
to unset it
array[i] ^= setbm1
Otherwise you can use http://java.sun.com/javase/6/docs/api/java/util/BitSet.html
