right shift in byte array

1
votes

I have two hex strings of 8 digits. I need to apply and operation for these two hex string, then apply right shift to 7 bits and get the decimal value. I have tried converting Hex strings to byte array of length 4 (8 *2 = 32 bits = 4 bytes) and did & operation to individual bytes in same order, saved the result to another byre array of length 4. How to do bit shifting to this byte array?

Ex : data1 in hex: 0x40003019,  
     data1 in bits: 0100-0000 0000-0000 0011-0000 0001-1001,
     data1 in bytes:    64   0  48  25,
     data2 in hex: 0x00FFFF80,
     data2 in bits : 0000-0000 1111-1111 1111-1111 1000-0000,
     data2 in bytes :  0 255 255 128

AND operation between data1Bytes , data2Bytes which gives output : bytearray1[0,0,48,0] (bits for these 0000-0000 0000-0000 0011-0000 0000-0000 and decimal value is 12,288).

Till this step all my conversions and calculations are working as expected. now I need to right shift 7 bits of this end result which should give 0000-0000 0000-0000 0000-0000 0110-0000( decimal value of 96).

1)I have tried converting byte array to int and apply right shift

 var res = BitConverter.ToInt32(bytearray1, 0);
 var shift = res >> 7;

but res = 3145728(which should be 12,228) and shift = 24,576(which should be 96)

2)I have tired converting bytearray1[0,0,48,0] into BitArray but bits in resultant BitArray are in reverse order 

var bitArray = new BitArray(bytearray1);

bitArray[0]...bitArray[19] = false, bitArray[20] = bitArray[21] = true , bitArray[22]...bitArray[31] = false.

bitArray[0] -----------[31] : 0000 0000 0000 0000 0000 1100 0000 0000,

bit shifting this result wrong value. Please help me with this, what I am missing?

2
c#.net
You say "right-shift" but your example left-shifts from 00110000 to 01100000. Which is correct? Perhaps you just need to replace >> with <<... – Matthew Watson
@MatthewWatson hi, if you are seeing last bit representation, I have just showed the bits stored in bit array after converting bytearray1 to bitArray. I need right shift my end result after doing and operation between bytes. – Naveen kumar
Hmm but you say now I need to right shift 7 bits of this end result which should give 0000-0000 0000-0000 0000-0000 **0110-0000** which, given that the previous step had 0000-0000 0000-0000 **0011-0000** 0000-0000, is a left shift. – Matthew Watson
Ah I think I see what you're doing now - it's a shift across multiple bytes. – Matthew Watson
@MatthewWatson Yes. you are right, its a shift across 4 bytes – Naveen kumar

2 Answers

0
votes

I'm not sure why this isn't working for you, but the obvious approach works when I try it.

Firstly assume you have the two hex numbers in uint values:

uint data1 = 0x40003019;
uint data2 = 0x00FFFF80;

Now just AND them together and then right shift the result:

uint anded  = data1 & data2;
uint result = anded >> 7; // Gives 96 as requested.

This gives a result of 96.

If your input is a string of the form string str = "0x40003019"; you can convert it to a uint like so:

uint data1 = uint.Parse(str.Substring(2), NumberStyles.HexNumber);

The str.SubString(2) is just to strip off the "0x" prefix. This is unnecessary if the input string does not have a "0x" prefix.

-2
votes

Try following : 

using System;
using System.Collections.Generic;
using System.Linq;
using System.Text;
using System.Xml;
using System.Xml.Linq;


namespace ConsoleApplication33
{

    class Program
    {

        static void Main(string[] args)
        {
            List<string> digits = new List<string>() {
                                  "0000","0001","0010","0011","0100","0101","0110","0111",
                                  "1000","1001","1010","1011","1100","1101","1110","1111"
                              };
            string input = "0100-0000 0000-0000 0011-0000 0001-1001";
            byte[] bytes = input.Split(new char[] { '-', ' ' }).Select(x => (byte)digits.IndexOf(x)).ToArray();
            ulong number = BitConverter.ToUInt64(bytes,0);
            Console.WriteLine(number);
            Console.ReadLine();
        }
    }


}