Help with bit wise opetation to get two values from a hex#

Miguel Martorell

Hi,
I have a program where I have to get two values from a HEX#

For ex. BE which in binary is equal to 10111110

Now the structure to get the two values is this:
The last 3 bits is one value.

So 110 = 6 <---I got this.

Here comes my problem.

The first 5 bits is a 5 bit signed integer and I can't seem to get it to work in c#.

The 5 bits in this example are 10111
On paper what I do it do the 1's complement and add 1. This has given me the value I need (the correct value).

Regular value = 10111
1's Comp = 01000
Add 1 = 01001 <---- = -9 since the first bit is a 1 in the regular value which makes it negative

My problem is transalting this in C#. I Changed the value to an int which would be 23.

Lets say a = 23.
How do I get from here to -9? The ~a gives me -24 and obviosuly adding 1 will give -23. Can anyone help? If you have any questions, dont hesitate to ask. Hopefully this was clear.

11 Replies

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Vulpes15y ago
Accepted answer
It's difficult to do something like this using normal bit arithmetic. I'd suggest the following instead:

using System;

class Test
{
   static void Main()
   {
   string s = "10111110";
   byte BE = Convert.ToByte(s, 2);
   int a;
   int b;
   ProcessByte(BE, out a, out b);
   Console.WriteLine("a is {0}", a);
   Console.WriteLine("b is {0}", b);
   Console.ReadKey();
   }

   static void ProcessByte(byte BE, out int a, out int b)
   {
   b = BE & 7;

   if (BE < 128)
   {
   a = (BE >> 3) & 31;
   }
   else
   {
   char[] bits = Convert.ToString(BE, 2).Substring(0, 5).ToCharArray();
   for(int i = 0; i < 5 ; i++) bits[i] = (bits[i] == '0') ? '1' : '0'; // flip bits
   a = Convert.ToInt32(new string(bits), 2);
   a++; // add one
   a = -a; // negate
   }
   }
}
Miguel Martorell15y ago
Vulpes: It's still better than me and you got the answer I was looking for. So it's good enough! Thank you!

Sam Hobbs : I'm sorry about that I knew it could be confusing or not have the wole information. That's why I explained it in more detail on the second post.
Still I appreaciate the help. Next time i'll try to give a more comprehensive explanation of my problem

Thank you all and have a great day.
Sam Hobbs15y ago
Miguel you did not give all the information in the initial question so I wasted my time trying to help.

Note that it is more important to describe the problem than it is to describe your guess at a solution.
Vulpes15y ago
Actually, Miguel, it's just struck me that there is an easy way to do it using ordinary bit arithmetic. All we need to do is to flip all the bits and then mask off the last 5 for a second time:

static void ProcessByte(byte BE, out int a, out int b)
   {
   b = BE & 7;
   a = (BE >> 3) & 31;

   if (BE > 127)
   {
   a = ~a & 31; // flip bits and remask last 5
   a++; // add one
   a = -a; // negate
   }
   }

Not as hot on this stuff as I once was :(
Miguel Martorell15y ago
Ohh duh, easy! (Alwasy easy when you understand it...jeje) That makes perfect sense. Thank you very much.
Now I can use this logic in other operations.
Take care and thanks once again for the help.
Vulpes15y ago
Well, 7 in binary is 111 and if you '&' a byte with 111 it masks off the last 3 bits.

Similarly, 31 in binary is 11111 and, if you shift the bits 3 places to the right (with >> 3) and then '&' with 11111 it masks off the first 5 bits of the byte.

If the byte value is less than 128, then we know that the leftmost bit is not set and so it can't be negative.

The ~ operator does in fact flip the bits of a byte but the trouble is that it flips all 8 (including the first three zero bits) and we only want to flip 5 here and leave the rest alone. I thought it would therefore be easier to do this part 'manually'.
Miguel Martorell15y ago
Well this worked very well.
Thank you very much.

By any chance can you explain a few things?

I can't firugre out how you came up with the &7 to get int a.
I mean I just did is take the hex substring (5,3) and converted to int and it worked but that seems easier. Tried out logic but not working.jeje . I want to know for in case others ask on why I took that option.

And the same goes with a = (BE >> 3) & 31; when it's less that 128.
How did u come up with it.

The only way to inverse the bits id with a for function. What exactly does the Not (~) operation do? Thsi si where i was getting stuck. Should have tried the shifting bits manually like you did.

I really appreaciate the help. This worked very well.
Sam Hobbs15y ago
Okay if a byte consists of 6 bits then another 6 bits then 4 bits then you can use a combination of shift rights and bitwise ands to get the data. Do you understand about shifting? It is getting late for me so if you need more clarification later then someone else can help you.
Sam Hobbs15y ago
So you want something such as the following.
Int32 a = 23;
Int32 n = a &0xF;
if ((a & 0x10) == 0x10)
n -= 16;
Miguel Martorell15y ago
The 5 bits is signed. so the first bit will just act as the sign and the rest would be the number. so it would go from -16 to +15. It could be negataive or positive, just used BE as an example.

The data that is read for example is 090B 35BE FF00 (hex).

09 is a code which is 09 in dec
0 is a parameter which is 0 dec
B is another parameter which is 11 in dec
3 is another parameter which is 3 in dec
5 is another parameter which is 5 in dec

BE conatains two parameters. To get the you convert to binary and the last 3 bits is one data (goes from 0 - 7)
and the first 5 bits is a signed int that goes from -16 to +15.

Now, in the first post i covered how in paper i get to the answer. I just dont get how to get that part in c#. I got all the rest of the values by converting hex to int.
but that part is different. (twos complement i believe)...if you need more examples id be glad to put more

Thank you for your help.
Sam Hobbs15y ago
I think there is not enough information about the format of the data.

Use the Windows Calculator accessory to try things out. Change the Calculator's view to Programmer. Put in the value 23. Click the radio button for Hex; you will see 17. Then click back to Dec. Click the "+-" button to make it negative. Then convert that to Hex; you should get FFFFFFFFFFFFFFE9. Do you understand why?

So I am not sure what the input data is. I am not sure whether there is a sign bit; since you say the output could be or should be negative, I am not sure about the sign bit.