# Tutorial :Comparing set of bits in byte array

### Question:

I have a byte array, as follows:

`byte[] array = new byte[] { 0xAB, 0x7B, 0xF0, 0xEA, 0x04, 0x2E, 0xF3, 0xA9};`

The task is to find the quantity of occurrences '0xA' in it. Could you advise what to do? The answer is 6.

### Solution:1

So from your comment, you want the total count of appearances of the bit pattern `1010` in the bytes in your array.

For a given byte `b`, the count is the sum of

``(b & 0x0A) == 0x0A ? 1 : 0  (b & 0x14) == 0x14 ? 1 : 0  (b & 0x28) == 0x28 ? 1 : 0  (b & 0x50) == 0x50 ? 1 : 0  (b & 0xA0) == 0xA0 ? 1 : 0  ``

(left as an exercise: what is this doing?)

Put this in a function, call it for each byte in the array, sum the results.

### Solution:2

If you treat the entire array as a single bit-string:

``0xAB,    0x7B,    0xF0,    0xEA,    0x04,    0x2E,    0xF3,    0xA9 is then:  10101011 01111011 11110000 11101010 00000100 00101110 11110011 10101001  ====                         ====                              ====    ====                         ====                              ====  ``

This has 1010 occurring 6 times.

If you don't try to match across byte boundaries, you could try something like the following (tested in Perl and translated by hand):

``int counter = 0;  for (int i = 0; i < array.length; ++i)  {      for (int bits = 0xA0, mask = 0xF0; bits >= 0x0A; bits >>= 1, mask >>= 1)      {          if ((array[i] & mask) == bits)              ++counter;      }  }  ``

To match across byte boundaries, you have to shift the bits in from the next byte. Try something like this (tested in Perl and translated by hand):

``int counter = 0;  byte tester = array[0];    for (int i = 1; i < array.length + 1; ++i)  {      byte nextByte = i < array.length ? array[i] : 0;        for (int bit = 0; bit < 8; ++bit)      {          if ((tester & 0xF0) == 0xA0)              ++counter;            tester <<= 1;          if ((nextByte & 0x80) != 0)              tester |= 1;            nextByte <<= 1;      }  }  ``

Both programs count 6 as there are no 1010 sequences across byte-boundaries in this example.

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