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How do I inverse this bitwise operation?

I have the below code and I can't understand why the last line doesn't return 77594624. Can anyone help me write the inverse bitwise operation to go from 77594624 to 4 and back to 77594624?
Console.WriteLine(77594624);
Console.WriteLine((77594624 >> 24) & 0x1F);
Console.WriteLine((4 & 0x1F) << 24);

When you bit shift a value you might "lose" bits during that operation. If you right shift the value 16, which is 0b10000 in binary, by 4, you will get 1.
0b10000 = 16
0b00001 = 1
But this is also the case for other numbers like 28, which is 0b11100.
0b11100 = 28 = 16 + 8 + 4
0b00001 = 1
So with the start point 1 you cannot go "back" to the original number by left shifting again, as there is not enough information/bits available, you don't know if you want to go back to 16 or 28.

77594624 looks like this in binary, and the x's mark the part that is extracted by the right shift and bitwise AND:
000001000101000000000000000000000
xxxxx
Clearly some information is lost.
If the other parts of the number were available as well, then they could be reassembled.

Convert 11 bit hex value to a signed 32 bit int

I am given a 11 bit signed hex value that must be stored in an int32 data type. When I cast the hex value to an int 32, the 11 bit hex value is obviously smaller then the int32 so it 0 fills the higher order bits.
Basically i need to be able to store 11 bit signed values in an int32 or 16 from a given 11 bit hex value.
For example.
String hex = 0x7FF;
If i cast this to int32 using Int.parse(hex, System.Globalization.Numbers.Hexvalue);
I get 2047 when it should be -1 (according to the 11 bit binary 111 1111)
How can I accomplish this in c#?

It's actually very simple, just two shifts. Shifting right keeps the sign, so that's useful. In order to use it, the sign of the 11 bit thing has to be aligned with the sign of the int:
x <<= -11;
Then do the right shift:
x >>= -11;
That's all.
The -11, which may seem odd, is just a shorter way to write 32 - 11. That's not in general the same thing, but shift counts are masked by 31 (ints) or 63 (longs), so in this case you can use that shortcut.

string hex = "0x7FF";
var i = (Convert.ToInt32(hex, 16) << 21) >> 21;

Reading 4 bits without losing information

I have come across a problem I cannot seem to solve.
I have a type of file "ASDF", and in their header I can get the necessary information to read them. The problem is that one of the "fields" is only 4 bits long.
So, lets say it's like this:
From bit 0 to 8 it's the index of the current node (I've read this already)
From 8 to 16 it's the index for the next node (Read this as well)
From bit 16 to 20 Length of the content (string, etc..)
So my problem is that if I try to read the "length" with a bytereader I will be losing 4 bits of information, or would be "4 bits off". Is there any way to read only 4 bits?

You should read this byte as you read the others then apply a bitmask of 0x0F
For example
byte result = (byte)(byteRead & 0x0F);
this will preserve the lower four bits in the result.
if the needed bits are the high four then you could apply the shift operator
byte result = (byte)((byteRead & 0x0F) >> 5);

Ignoring Leftmost Bit in Four Bytes

First an explanation of why:
I have a list of links to a variety of MP3 links and I'm trying to read the ID3 information for these files quickly. I'm only downloading the first 1500 or so bytes and trying to ana yze the data within this chunk. I came across ID3Lib, but I could only get it to work on completely downloaded files and didn't notice any support for Streams. (If I'm wrong in this, feel free to point that out)
So basically, I'm left trying to parse the ID3 tag by myself. The size of the tag can be determined from four bytes near the start of the file. From the ID3 site:
The ID3v2 tag size is encoded with four bytes where the most
significant bit (bit 7) is set to zero in every byte, making a total
of 28 bits. The zeroed bits are ignored, so a 257 bytes long tag is
represented as $00 00 02 01.
So basically:
00000000 00000000 00000010 00000001
becomes
0000 00000000 00000001 00000001
I'm not too familiar with bit level operations and was wondering if someone could shed some insight on an elegant solution to ignore the leftmost bit of each of these four bytes? I'm trying to pull a base 10 integer from it, so that works as well.

If you've got the four individual bytes, you'd want:
int value = ((byte1 & 0x7f) << 21) |
((byte2 & 0x7f) << 14) |
((byte3 & 0x7f) << 7) |
((byte4 & 0x7f) << 0);
If you've got it in a single int already:
int value = ((value & 0x7f000000) >> 3) |
((value & 0x7f0000) >> 2) |
((value & 0x7f00) >> 1) |
(value & 0x7f);

To clear the most significant bit, AND with 127 (0x7F), this takes all bits apart from the MSB.
int tag1 = tag1Byte & 0x7F; // this is the first one read from the file
int tag2 = tag2Byte & 0x7F;
int tag3 = tag3Byte & 0x7F;
int tag4 = tag4Byte & 0x7F; // this is the last one
To convert this into a single number, realize that each tag value is a base 128 digit. So, the least signifiant is multipled by 128^0 (1), the next 128^1 (128), the third significant (128^2) and so on.
int tagLength = tag4+(tag3<<7)+(tag2<<14)+(tag1<<21)
You mention you want to conver this to base 10. You can then convert this to base 10, say for printing, using int to string conversion:
String base10 = String.valueOf(tagLength);

Why AND two numbers to get a Boolean?

I am working on a little Hardware interface project based on the Velleman k8055 board.
The example code comes in VB.Net and I'm rewriting this into C#, mostly to have a chance to step through the code and make sense of it all.
One thing has me baffled though:
At one stage they read all digital inputs and then set a checkbox based on the answer to the read digital inputs (which come back in an Integer) and then they AND this with a number:
i = ReadAllDigital
cbi(1).Checked = (i And 1)
cbi(2).Checked = (i And 2) \ 2
cbi(3).Checked = (i And 4) \ 4
cbi(4).Checked = (i And 8) \ 8
cbi(5).Checked = (i And 16) \ 16
I have not done Digital systems in a while and I understand what they are trying to do but what effect would it have to AND two numbers? Doesn't everything above 0 equate to true?
How would you translate this to C#?

This is doing a bitwise AND, not a logical AND.
Each of those basically determines whether a single bit in i is set, for instance:
5 AND 4 = 4
5 AND 2 = 0
5 AND 1 = 1
(Because 5 = binary 101, and 4, 2 and 1 are the decimal values of binary 100, 010 and 001 respectively.)

I think you 'll have to translate it to this:
i & 1 == 1
i & 2 == 2
i & 4 == 4
etc...
This is using the bitwise AND operator.
When you use the bitwise AND operator, this operator will compare the binary representation of the two given values, and return a binary value where only those bits are set, that are also set in the two operands.
For instance, when you do this:
2 & 2
It will do this:
0010 & 0010
And this will result in:
0010
0010
&----
0010
Then if you compare this result with 2 (0010), it will ofcourse return true.

Just to add:
It's called bitmasking
http://en.wikipedia.org/wiki/Mask_(computing)
A boolean only require 1 bit. In the implementation most programming language, a boolean takes more than a single bit. In PC this won't be a big waste, but embedded system usually have very limited memory space, so the waste is really significant. To save space, the booleans are packed together, this way a boolean variable only takes up 1 bit.
You can think of it as doing something like an array indexing operation, with a byte (= 8 bits) becoming like an array of 8 boolean variables, so maybe that's your answer: use an array of booleans.

Think of this in binary e.g.
10101010
AND
00000010
yields 00000010
i.e. not zero. Now if the first value was
10101000
you'd get
00000000
i.e. zero.
Note the further division to reduce everything to 1 or 0.

(i and 16) / 16 extracts the value (1 or 0) of the 5th bit.
1xxxx and 16 = 16 / 16 = 1
0xxxx and 16 = 0 / 16 = 0

And operator performs "...bitwise conjunction on two numeric expressions", which maps to '|' in C#. The '` is an integer division, and equivalent in C# is /, provided that both operands are integer types.

The constant numbers are masks (think of them in binary). So what the code does is apply the bitwise AND operator on the byte and the mask and divide by the number, in order to get the bit.
For example:
xxxxxxxx & 00000100 = 00000x000
if x == 1
00000x00 / 00000100 = 000000001
else if x == 0
00000x00 / 00000100 = 000000000

In C# use the BitArray class to directly index individual bits.
To set an individual bit i is straightforward:
b |= 1 << i;
To reset an individual bit i is a little more awkward:
b &= ~(1 << i);
Be aware that both the bitwise operators and the shift operators tend to promote everything to int which may unexpectedly require casting.

As said this is a bitwise AND, not a logical AND. I do see that this has been said quite a few times before me, but IMO the explanations are not so easy to understand.
I like to think of it like this:
Write up the binary numbers under each other (here I'm doing 5 and 1):
101
001
Now we need to turn this into a binary number, where all the 1's from the 1st number, that is also in the second one gets transfered, that is - in this case:
001
In this case we see it gives the same number as the 2nd number, in which this operation (in VB) returns true. Let's look at the other examples (using 5 as i):
(5 and 2)
101
010
----
000
(false)
(5 and 4)
101
100
---
100
(true)
(5 and 8)
0101
1000
----
0000
(false)
(5 and 16)
00101
10000
-----
00000
(false)
EDIT: and obviously I miss the entire point of the question - here's the translation to C#:
cbi[1].Checked = i & 1 == 1;
cbi[2].Checked = i & 2 == 2;
cbi[3].Checked = i & 4 == 4;
cbi[4].Checked = i & 8 == 8;
cbi[5].Checked = i & 16 == 16;

I prefer to use hexadecimal notation when bit twiddling (e.g. 0x10 instead of 16). It makes more sense as you increase your bit depths as 0x20000 is better than 131072.