C＃对十六进制值的二进制运算

Question

I have a hex value that eg following octets: 82 A9 43

I need to remove the 8 bit of all of them and concatenate them together.

eg: if we change each to binary:

82 = 10000010
A9 = 10101001
43 = 01000011

Now I need to remove the MSB on all of the above with the following code:

                    int t = 0x7F;
                    int hv1 = 0x82;
                    int hv2 = 0xA9;
                    int hv3 = 0x43;

                    int r1 = hv1 & t;
                    int r2 = hv2 & t;
                    int r3 = hv3 & t;

After the above procedure it looks like:

r1 = 10
r2 = 101001
r3 = 100011

But I need as follows:

r1 = 0000010
r2 = 0101001
r3 = 0100011

The reason is later I wanted concatenate all the r1 | r2 | r3 to be as follows:

000001001010010100011

The above should then be converted to integer.

Do you have a clue, how can this be done. Thank you for your help.

Answer 1

Leading 0's on binary values don't really do anything, so you can just count the characters and add the 0's back to the front of them, and then concatenate them together

string PadLeft (string s) {
   return string.Format("({0,8})", s)
}

Answer 2

It is not clear how you are converting your integer values into the binary strings, but I imagine you are doing something like this:

string binaryR1 = Convert.ToString(r1, 2);//binaryR1 = "10"

You can then pad the string using the PadLeft() function:

binaryR1 = binaryR1.PadLeft(7, '0');//binaryR1 = "0000010"

or in a single line:

string binaryR1 = Convert.ToString(r1, 2).PadLeft(7, '0');//binaryR1 = "0000010"

NOTE: These values have been padded to 7 digits as you do not require the MSB

Applying this to all 3 values would then allow you to concat into the full string:

string full = binaryR1 + binaryR2 + binaryR3;//full = "000001001010010100011"

Answer 3

There is no need to convert to strings then back to integers, just do integer math like this:

int concat = r1 << 16 | r2 << 8 | r3;

or more briefly:

int ttt = 0x7f7f7f; 
int concat = (hv1 << 16 | hv2 << 8 | hv3 & ttt;

Answer 4

Numerically, leading zeroes are irrelevant. Have your function return a String representation of the number so you can pad it to 8 characters.

Answer 5

if you want them as you've described them, just tack on some zero's to the front, it won't change the binary value

private static string AddZeros(string r)
    {
        while (r.Length < 8)
            r = "0" + r;

        return r;
    }

Answer 6

Try something like this:

int r1 = hv1 & 0x7F;
int r2 = hv2 & 0x7F;
int r3 = hv3 & 0x7F;
int res = ( r1 << 14 ) | ( r2 << 7 ) | r3;

That should strip the 8th bit of each byte, then concatenate bit 0-7 of each into a 21-bit number:

111111122222223333333 // where 1 are bits from r1, 2 from r2 and so on

Is that what you wanted?