Hello!
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.
Hello!
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.
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.
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.