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Efficiently shift an integer to the left and wrap around [duplicate]

I'm trying to implement a function that performs a circular rotation of a byte to the left and to the right.
I wrote the same code for both operations. For example, if you are rotating left 1010 becomes 0101. Is this right?
unsigned char rotl(unsigned char c) {
int w;
unsigned char s = c;
for (w = 7; w >= 0; w--) {
int b = (int)getBit(c, w);//
if (b == 0) {
s = clearBit(s, 7 - w);
} else if (b == 1) {
s = setBit(s, 7 - w);
}
}
return s;
}
unsigned char getBit(unsigned char c, int n) {
return c = (c & (1 << n)) >> n;
}
unsigned char setBit(unsigned char c, int n) {
return c = c | (1 << n);
}
unsigned char clearBit(unsigned char c, int n) {
return c = c &(~(1 << n));
}

There is no rotation operator in C, but if you write:
unsigned char rotl(unsigned char c)
{
return (c << 1) | (c >> 7);
}
then, according to this: http://www.linux-kongress.org/2009/slides/compiler_survey_felix_von_leitner.pdf (page 56), compilers will figure out what you want to do and perform the rotation it in only one (very fast) instruction.

Reading the answers and comments so far, there seems to be some confusion about what you are trying to accomplish - this may be because of the words you use. In bit manipulation, there are several "standard" things you can do. I will summarize some of these to help clarify different concepts. In all that follows, I will use abcdefgh to denote 8 bits (could be ones or zeros) - and as they move around, the same letter will refer to the same bit (maybe in a different position); if a bit becomes "definitely 0 or 1, I will denote it as such).
1) Bit shifting: This is essentially a "fast multiply or divide by a power of 2". The symbol used is << for "left shift" (multiply) or >> for right shift (divide). Thus
abcdefgh >> 2 = 00abcdef
(equivalent to "divide by four") and
abcdefgh << 3 = abcdefgh000
(equivalent to "multiply by eight" - and assuming there was "space" to shift the abc into; otherwise this might result in an overflow)
2) Bit masking: sometimes you want to set certain bits to zero. You do this by doing an AND operation with a number that has ones where you want to preserve a bit, and zeros where you want to clear a bit.
abcdefgh & 01011010 = 0b0de0g0
Or if you want to make sure certain bits are one, you use the OR operation:
abcdefgh | 01011010 = a1c11f1h
3) Circular shift: this is a bit trickier - there are instances where you want to "move bits around", with the ones that "fall off at one end" re-appearing at the other end. There is no symbol for this in C, and no "quick instruction" (although most processors have a built-in instruction which assembler code can take advantage of for FFT calculations and such). If you want to do a "left circular shift" by three positions:
circshift(abcdefgh, 3) = defghabc
(note: there is no circshift function in the standard C libraries, although it exists in other languages - e.g. Matlab). By the same token a "right shift" would be
circshift(abcdefgh, -2) = ghabcdef
4) Bit reversal: Sometimes you need to reverse the bits in a number. When reversing the bits, there is no "left" or "right" - reversed is reversed:
reverse(abcdefgh) = hgfedcba
Again, there isn't actually a "reverse" function in standard C libraries.
Now, let's take a look at some tricks for implementing these last two functions (circshift and reverse) in C. There are entire websites devoted to "clever ways to manipulate bits" - see for example this excellent one. for a wonderful collection of "bit hacks", although some of these may be a little advanced...
unsigned char circshift(unsigned char x, int n) {
return (x << n) | (x >> (8 - n));
}
This uses two tricks from the above: shifting bits, and using the OR operation to set bits to specific values. Let's look at how it works, for n = 3 (note - I am ignoring bits above the 8th bit since the return type of the function is unsigned char):
(abcdefgh << 3) = defgh000
(abcdefgh >> (8 - 3)) = 00000abc
Taking the bitwise OR of these two gives
defgh000 | 00000abc = defghabc
Which is exactly the result we wanted. Note also that a << n is the same as a >> (-n); in other words, right shifting by a negative number is the same as left shifting by a positive number, and vice versa.
Now let's look at the reverse function. There are "fast ways" and "slow ways" to do this. Your code above gave a "very slow" way - let me show you a "very fast" way, assuming that your compiler allows the use of 64 bit (long long) integers.
unsigned char reverse(unsigned char b) {
return (b * 0x0202020202ULL & 0x010884422010ULL) % 1023;
}
You may ask yourself "what just happened"??? Let me show you:
b = abcdefgh
* 0x0000000202020202 = 00000000 00000000 0000000a bcdefgha bcdefgha bcdefgha bcdefgha bcdefgh0
& 0x0000010884422010 = 00000000 00000000 00000001 00001000 10000100 01000010 00100000 00010000
= 00000000 00000000 0000000a 0000f000 b0000g00 0c0000h0 00d00000 000e0000
Note that we now have all the bits exactly once - they are just in a rather strange pattern. The modulo 1023 division "collapses" the bits of interest on top of each other - it's like magic, and I can't explain it. The result is indeed
hgfedcba
A slightly less obscure way to achieve the same thing (less efficient, but works for larger numbers quite efficiently) recognizes that if you swap adjacent bits , then adjacent bit pairs, then adjacent nibbles (4 bit groups), etc - you end up with a complete bit reversal. In that case, a byte reversal becomes
unsigned char bytereverse(unsigned char b) {
b = (b & 0x55) << 1 | (b & 0xAA) >> 1; // swap adjacent bits
b = (b & 0x33) << 2 | (b & 0xCC) >> 2; // swap adjacent pairs
b = (b & 0x0F) << 4 | (b & 0xF0) >> 4; // swap nibbles
return b;
}
In this case the following happens to byte b = abcdefgh:
b & 0x55 = abcdefgh & 01010101 = 0b0d0f0h << 1 = b0d0f0h0
b & 0xAA = abcdefgh & 10101010 = a0c0e0g0 >> 1 = 0a0c0e0g
OR these two to get badcfehg
Next line:
b & 0x33 = badcfehg & 00110011 = 00dc00hg << 2 = dc00hg00
b & 0xCC = badcfehg & 11001100 = ba00fe00 >> 2 = 00ba00fe
OR these to get dcbahgfe
last line:
b & 0x0F = dcbahgfe & 00001111 = 0000hgfe << 4 = hgfe0000
b & 0xF0 = dcbahgfe & 11110000 = dcba0000 >> 4 = 0000dcba
OR these to get hgfedcba
Which is the reversed byte you were after. It should be easy to see how just a couple more lines (similar to the above) get you to a reversed integer (32 bits). As the size of the number increases, this trick becomes more and more efficient, comparatively.
I trust that the answer you were looking for is "somewhere" in the above. If nothing else I hope you have a clearer understanding of the possibilities of bit manipulation in C.

If, as according to your comments, you want to shift one bit exactly, then one easy way to accomplish that would be this:
unsigned char rotl(unsigned char c)
{
return((c << 1) | (c >> 7));
}
What your code does is reversing the bits; not rotating them. For instance, it would make 10111001 into 10011101, not 01110011.

Pascal to C# conversion

I am trying to convert this pascal code into C# in order to communicate with a peripheral device attached to a comm port. This piece of code should calculate the Control Byte, however I'm not getting the right hex Value therefore I'm wondering if I'm converting the code in the right way.
Pascal:
begin
check := 255;
for i:= 3 to length(sequence)-4 do
check := check xor byte(sequence[i]);
end;
C#:
int check = 255;
for (int x = 3; x < (sequence.Length - 4); x++)
{
check = check ^ (byte)(sequence[x]);
}
Pascal function:
{ *** conversion of number into string ‘hex’ *** }
function word_to_hex (w: word) : string;
var
i : integer;
s : string;
b : byte;
c : char;
begin
s := ‘’;
for i:= 0 to 3 do
begin
b := (hi(w) shr 4) and 15;
case b of
0..9 : c := char(b+$30);
10..15 : c := char(b+$41-10);
end;
s := s + c;
w := w shl 4;
end;
word_ to_hex := s;
end;
C# Equivalent:
public string ControlByte(string check)
{
string s = "";
byte b;
char c = '\0';
//shift = check >> 4 & 15;
for (int x = 0; x <= 3; x++)
{
b = (byte)((Convert.ToInt32(check) >> 4) & 15);
if (b >= 0 && b <= 9)
{
c = (char)(b + 0x30);
}
else if (b >= 10 && b <= 15)
{
c = (char)(b + 0x41 - 10);
}
s = s + c;
check = (Convert.ToInt32(check) << 4).ToString();
}
return s;
}
And last pascal:
function byte_to_hex (b:byte) : string;
begin
byte_to_hex := copy(word_to_hex(word(b)),3,2);
end;
which i am not sure how is substringing the result from the function. So please let me know if there is something wrong with the code conversion and whether I need to convert the function result into bytes. I appreciate your help, UF.
Further info EDIT: Initially I send a string sequence containing the command and information that printer is supposed to print. Since every sequence has a unique Control Byte (in Hex) I have to calculate this from the sequence (sequence = "P1;1$l201PrinterPrinterPrinter1B/100.00/100.00/0/\") which is what upper code does according to POSNET=>"cc – control byte, encoded as 2 HEX digits (EXOR of all characters after ESC P to this byte with #255 initial quantity), according to the following algorithm in PASCAL language:(see first code block)".=>1. check number calculated in the above loop which constitutes control byte should be recoded into two HEX characters (ASCII characters from scope: ‘0’..’9’,’A’..’F’,’a’..’f’), utilizing the following byte_to_hex function:(see third code block). =>{* conversion of byte into 2 characters *}(see 5th code block)

The most obvious problem that I can see is that the Pascal code operates on 1-based 8 bit encoded strings, and the C# code operates on 0-based 16 bit encoded strings. To convert the Pascal/Delphi code that you use to C# you need to address the mis-match. Perhaps like this:
byte[] bytes = Encoding.Default.GetBytes(sequence);
int check = 255;
for (int i = 2; i < bytes.Length-4; i++)
{
check ^= bytes[i];
}
Now, in order to write this I've had to make quite a few assumptions, because you did not include anywhere near enough code in the question. Here's what I assumed:
The Pascal sequence variable is a 1-based 8 bit ANSI encoded Delphi AnsiString.
The Pascal check variable is a Delphi 32 bit signed Integer.
The C# sequence variable is a C# string.
If any of those assumptions prove to be false, then the code above will be no good. For instance, perhaps the Pascal check is really Byte. In which case I guess the C# code should be:
byte[] bytes = Encoding.Default.GetBytes(sequence);
byte check = 255;
for (int i = 2; i < bytes.Length - 4; i++)
{
check ^= bytes[i];
}
I hope that this persuades you of the importance of supplying complete information.
That's really all the meat of this question. The rest of the code concerns converting values to hex strings in C# code. That has been covered again and again here on Stack Overflow. For instance:
C# convert integer to hex and back again
How do you convert Byte Array to Hexadecimal String, and vice versa?
There are many many more such questions.

Microsoft BigInteger goes "negative" when I import from an array

Suppose I have an array that ends with the most signifiant bit of the most significant byte equaling 1.
My understanding is that if this is the case, BigInteger will treat this as a negative number, by design.
BigInteger numberToShorten = new BigInteger(toEncode);
if (numberToShorten.Sign == -1)
{
// problem with twos compliment or the last bit of the last byte is equal to 1
throw new Exception("Unexpected negative number");
}
To solve this problem, I think I need to add a dummy zero bit to the array, prior to converting the array. I can easily do this using Array.Resize().
My question is, how should I test if the last bit is indeed equal to 1?
I'm pretty weak on my boolean logic now, and am thinking I need to AND the values and test for equality, but not able to get the syntax correct in C#. Something like this:
byte temp = toEncode[toEncode.Length - 1];
if (temp == ???)
{
Array.Resize(ref toEncode, toEncode.Length +1);
}

If the number to be converted to BigInteger is always positive then you don't actually need to test at all; just appending a zero byte will always work correctly.
For completeness, checking if the MSB of any one byte is set is done with
if (byte & 0x80 == 0x80) ...

Found the answer at the bottom of this MSDN article (I think)
ulong originalNumber = UInt64.MaxValue;
byte[] bytes = BitConverter.GetBytes(originalNumber);
if (originalNumber > 0 && (bytes[bytes.Length - 1] & 0x80) > 0)
{
byte[] temp = new byte[bytes.Length];
Array.Copy(bytes, temp, bytes.Length);
bytes = new byte[temp.Length + 1];
Array.Copy(temp, bytes, temp.Length);
}
BigInteger newNumber = new BigInteger(bytes);
Console.WriteLine("Converted the UInt64 value {0:N0} to {1:N0}.",
originalNumber, newNumber);
// The example displays the following output:
// Converted the UInt64 value 18,446,744,073,709,551,615 to 18,446,744,073,709,551,615.

I'm not sure if this is necessary, have you tried it? In any case, assuming that temp contains a byte and you are trying to test whether that byte ends in 1, this is how you should do it:
if((temp & 0x01)==1)
{
...
}
That amounts to:
Temp: ???????z
0x01: 00000001
Temp & 0x01: 0000000z
Where z represents the bit you are trying to test, and you can now just test whether the result is equal to 1 or not.
If you are trying to test whether the byte starts with 1, do:
if((temp & 0x80)==0x80) //0x80 is 10000000 in binary
{
...
}

How do I properly loop through and print bits of an Int, Long, Float, or BigInteger?

I'm trying to debug some bit shifting operations and I need to visualize the bits as they exist before and after a Bit-Shifting operation.
I read from this answer that I may need to handle backfill from the shifting, but I'm not sure what that means.
I think that by asking this question (how do I print the bits in a int) I can figure out what the backfill is, and perhaps some other questions I have.
Here is my sample code so far.
static string GetBits(int num)
{
StringBuilder sb = new StringBuilder();
uint bits = (uint)num;
while (bits!=0)
{
bits >>= 1;
isBitSet = // somehow do an | operation on the first bit.
// I'm unsure if it's possible to handle different data types here
// or if unsafe code and a PTR is needed
if (isBitSet)
sb.Append("1");
else
sb.Append("0");
}
}

Convert.ToString(56,2).PadLeft(8,'0') returns "00111000"
This is for a byte, works for int also, just increase the numbers

To test if the last bit is set you could use:
isBitSet = ((bits & 1) == 1);
But you should do so before shifting right (not after), otherwise you's missing the first bit:
isBitSet = ((bits & 1) == 1);
bits = bits >> 1;
But a better option would be to use the static methods of the BitConverter class to get the actual bytes used to represent the number in memory into a byte array. The advantage (or disadvantage depending on your needs) of this method is that this reflects the endianness of the machine running the code.
byte[] bytes = BitConverter.GetBytes(num);
int bitPos = 0;
while(bitPos < 8 * bytes.Length)
{
int byteIndex = bitPos / 8;
int offset = bitPos % 8;
bool isSet = (bytes[byteIndex] & (1 << offset)) != 0;
// isSet = [True] if the bit at bitPos is set, false otherwise
bitPos++;
}

how to loop through the digits of a binary number?

I have a binary number 1011011, how can I loop through all these binary digits one after the other ?
I know how to do this for decimal integers by using modulo and division.

int n = 0x5b; // 1011011
Really you should just do this, hexadecimal in general is much better representation:
printf("%x", n); // this prints "5b"
To get it in binary, (with emphasis on easy understanding) try something like this:
printf("%s", "0b"); // common prefix to denote that binary follows
bool leading = true; // we're processing leading zeroes
// starting with the most significant bit to the least
for (int i = sizeof(n) * CHAR_BIT - 1; i >= 0; --i) {
int bit = (n >> i) & 1;
leading |= bit; // if the bit is 1, we are no longer reading leading zeroes
if (!leading)
printf("%d", bit);
}
if (leading) // all zero, so just print 0
printf("0");
// at this point, for n = 0x5b, we'll have printed 0b1011011

You can use modulo and division by 2 exactly like you would in base 10. You can also use binary operators, but if you already know how to do that in base 10, it would be easier if you just used division and modulo

Expanding on Frédéric and Gabi's answers, all you need to do is realise that the rules in base 2 are no different to in base 10 - you just need to do your division and modulus with a divisor 2 instead of 10.
The next step is simply to use number >> 1 instead of number / 2 and number & 0x1 instead of number % 2 to improve performance. Mind you, with modern optimising compilers there's probably no difference...

Use an AND with increasing powers of two...

In C, at least, you can do something like:
while (val != 0)
{
printf("%d", val&0x1);
val = val>>1;
}

To expand on #Marco's answer with an example:
uint value = 0x82fa9281;
for (int i = 0; i < 32; i++)
{
bool set = (value & 0x1) != 0;
value >>= 1;
Console.WriteLine("Bit set: {0}", set);
}
What this does is test the last bit, and then shift everything one bit.

If you're already starting with a string, you could just iterate through each of the characters in the string:
var values = "1011011".Reverse().ToCharArray();
for(var index = 0; index < values.Length; index++) {
var isSet = (Boolean)Int32.Parse(values[index]); // Boolean.Parse only works on "true"/"false", not 0/1
// do whatever
}

byte input = Convert.ToByte("1011011", 2);
BitArray arr = new BitArray(new[] { input });
foreach (bool value in arr)
{
// ...
}

You can simply loop through every bit. The following C like pseudocode allows you to set the bit number you want to check. (You might also want to google endianness)
for()
{
bitnumber = <your bit>
printf("%d",(val & 1<<bitnumber)?1:0);
}
The code basically writes 1 if the bit it set or 0 if not. We shift the value 1 (which in binary is 1 ;) ) the number of bits set in bitnumber and then we AND it with the value in val to see if it matches up. Simple as that!
So if bitnumber is 3 we simply do this
00000100 ( The value 1 is shifted 3 left for example)
AND
10110110 (We check it with whatever you're value is)
=
00000100 = True! - Both values have bit 3 set!