I have do not have much knowledge of C and I'm stuck with a problem since one of my colleague is on leave.
I have a 32 bit number and i have to extract bits from it. I did go through a few threads but I'm still not clear how to do so. I would be highly obliged if someone can help me.
Here is an example of what I need to do:
Assume hex number = 0xD7448EAB.
In binary = 1101 0111 0100 0100 1000 1110 1010 1011.
I need to extract the 16 bits, and output that value. I want bits 10 through 25.
The lower 10 bits (Decimal) are ignored. i.e., 10 1010 1011 are ignored.
And the upper 6 bits (Overflow) are ignored. i.e. 1101 01 are ignored.
The remaining 16 bits of data needs to be the output which is 11 0100 0100 1000 11 (numbers in italics are needed as the output).
This was an example but I will keep getting different hex numbers all the time and I need to extract the same bits as I explained.
How do I solve this?
Thank you.
For this example you would output 1101 0001 0010 0011, which is 0xD123, or 53,539 decimal.
You need masks to get the bits you want. Masks are numbers that you can use to sift through bits in the manner you want (keep bits, delete/clear bits, modify numbers etc). What you need to know are the AND, OR, XOR, NOT, and shifting operations. For what you need, you'll only need a couple.
You know shifting: x << y moves bits from x *y positions to the left*.
How to get x bits set to 1 in order: (1 << x) - 1
How to get x bits set to 1, in order, starting from y to y + x: ((1 << x) -1) << y
The above is your mask for the bits you need. So for example if you want 16 bits of 0xD7448EAB, from 10 to 25, you'll need the above, for x = 16 and y = 10.
And now to get the bits you want, just AND your number 0xD7448EAB with the mask above and you'll get the masked 0xD7448EAB with only the bits you want. Later, if you want to go through each one, you'll need to shift your result by 10 to the right and process each bit at a time (at position 0).
The answer may be a bit longer, but it's better design than just hard coding with 0xff or whatever.
OK, here's how I wrote it:
#include <stdint.h>
#include <stdio.h>
main() {
uint32_t in = 0xd7448eab;
uint16_t out = 0;
out = in >> 10; // Shift right 10 bits
out &= 0xffff; // Only lower 16 bits
printf("%x\n",out);
}
The in >> 10 shifts the number right 10 bits; the & 0xffff discards all bits except the lower 16 bits.
I want bits 10 through 25.
You can do this:
unsigned int number = 0xD7448EAB;
unsigned int value = (number & 0x3FFFC00) >> 10;
Or this:
unsigned int number = 0xD7448EAB;
unsigned int value = (number >> 10) & 0xFFFF;
I combined the top 2 answers above to write a C program that extracts the bits for any range of bits (not just 10 through 25) of a 32-bit unsigned int. The way the function works is that it returns bits lo to hi (inclusive) of num.
#include <stdio.h>
#include <stdint.h>
unsigned extract(unsigned num, unsigned hi, unsigned lo) {
uint32_t range = (hi - lo + 1); //number of bits to be extracted
//shifting a number by the number of bits it has produces inconsistent
//results across machines so we need a special case for extract(num, 31, 0)
if(range == 32)
return num;
uint32_t result = 0;
//following the rule above, ((1 << x) - 1) << y) makes the mask:
uint32_t mask = ((1 << range) -1) << lo;
//AND num and mask to get only the bits in our range
result = num & mask;
result = result >> lo; //gets rid of trailing 0s
return result;
}
int main() {
unsigned int num = 0xd7448eab;
printf("0x%x\n", extract(num, 10, 25));
}
Related
I'm working with bitshift for the first time and I'm experiencing unexpected results.
I'm declaring the shift amount as follows:
byte p_size = 0;
if (ver == 0x12 || ver == 0x13)
p_size = 20;
else
p_size = 40;
The value to be shifted is declared as
int t_size = rinput.ReadInt32();
And finally the code I use to shift:
int temp = t_size >> p_size << p_size;
Let's say t_size = 0x2000385E and p_size = 20. temp = 0x20000000 as expected.
Now if t_size = 0x40001014 and p_size = 40, temp = 0x40001000 instead of 0x40000000. I calculated "manually" using a bitwise calculator and it matches the expected result of 0x40000000.
It's probably a silly oversight on my behalf but I don't understand what would cause the weird results with p_size = 40... any advice is appreciated!
Shifting a 32 integer by 40 bits doesn't really make sense since you would be shifting the integer by more bits than it contains.
Both the left and right shift operators document what they do in this case:
If the first operand is an int or uint (32-bit quantity), the shift
count is given by the low-order five bits of the second operand
(second operand & 0x1f).
So when p_size is 40, the shifts are shifting by 40 & 0x1f = 8 bits.
If you need to shift by 40 bits, but your value into long.
Current behavior is expected as 40 & 0x1f is 8 as described in operator >>
If the first operand is an int or uint (32-bit quantity), the shift count is given by the low-order five bits of the second operand (second operand & 0x1f).
You probably looking for some masking rather than shifts - maybe
t_size & 0xFF000000
How to remove the leftmost bit?
I have a hexadecimal value BF
Its binary representation is 1011 1111
How can I remove the first bit, which is 1, and then it will become 0111 1110?
How to add "0" also to its last part?
To set bit N of variable x to 0
x &= ~(1 << N);
How it works: The expression 1 << N is one bit shifted N times to the left. For N = 7, this would be
1000 0000
The bitwise NOT operator ~ inverts this to
0111 1111
Then the result is bitwise ANDed with x, giving:
xxxx xxxx
0111 1111
--------- [AND]
0xxx xxxx
Result: bit 7 (zero-based count starting from the LSB) is turned off, all others retain their previous values.
To set bit N of variable x to 1
x |= 1 << N;
How it works: this time we take the shifted bit and bitwise OR it with x, giving:
xxxx xxxx
1000 0000
--------- [OR]
1xxx xxxx
Result: Bit 7 is turned on, all others retain their previous values.
Finding highest order bit set to 1:
If you don't know which is the highest bit set to 1 you can find out on the fly. There are many ways of doing this; a reasonable approach is
int x = 0xbf;
int highestSetBit = -1; // assume that to begin with, x is all zeroes
while (x != 0) {
++highestSetBit;
x >>= 1;
}
At the end of the loop, highestSetBit will be 7 as expected.
See it in action.
int i=0xbf;
int j=(i<<1) & 0xff;
or you could do:
(i*2) && 0xff
if you'd rather not do bit twiddling. >>1 is the equivalent of /2, and <<1 is the equivalent of *2.
I need to convert 16-bit XRGB1555 into 24-bit RGB888. My function for this is below, but it's not perfect, i.e. a value of 0b11111 wil give 248 as the pixel value, not 255. This function is for little-endian, but can easily be modified for big-endian.
public static Color XRGB1555(byte b0, byte b1)
{
return Color.FromArgb(0xFF, (b1 & 0x7C) << 1, ((b1 & 0x03) << 6) | ((b0 & 0xE0) >> 2), (b0 & 0x1F) << 3);
}
Any ideas how to make it work?
You would normally copy the highest bits down to the bottom bits, so if you had five bits as follows:
Bit position: 4 3 2 1 0
Bit variable: A B C D E
You would extend that to eight bits as:
Bit position: 7 6 5 4 3 2 1 0
Bit variable: A B C D E A B C
That way, all zeros remains all zeros, all ones becomes all ones, and values in between scale appropriately.
(Note that A,B,C etc aren't supposed to be hex digits - they are variables representing a single bit).
I'd go with a lookup table. Since there are only 32 different values it even fits in a cache-line.
You can get the 8 bit value from the 5 bit value with:
return (x<<3)||(x>>2);
The rounding might not be perfect though. I.e. the result isn't always closest to the input, but it never is further away that 1/255.
Assuming the 64bit integer 0x000000000000FFFF which would be represented as
00000000 00000000 00000000 00000000
00000000 00000000 >11111111 11111111
How do I find the amount of unset bits to the left of the most significant set bit (the one marked with >) ?
In straight C (long long are 64 bit on my setup), taken from similar Java implementations: (updated after a little more reading on Hamming weight)
A little more explanation: The top part just sets all bit to the right of the most significant 1, and then negates it. (i.e. all the 0's to the 'left' of the most significant 1 are now 1's and everything else is 0).
Then I used a Hamming Weight implementation to count the bits.
unsigned long long i = 0x0000000000000000LLU;
i |= i >> 1;
i |= i >> 2;
i |= i >> 4;
i |= i >> 8;
i |= i >> 16;
i |= i >> 32;
// Highest bit in input and all lower bits are now set. Invert to set the bits to count.
i=~i;
i -= (i >> 1) & 0x5555555555555555LLU; // each 2 bits now contains a count
i = (i & 0x3333333333333333LLU) + ((i >> 2) & 0x3333333333333333LLU); // each 4 bits now contains a count
i = (i + (i >> 4)) & 0x0f0f0f0f0f0f0f0fLLU; // each 8 bits now contains a count
i *= 0x0101010101010101LLU; // add each byte to all the bytes above it
i >>= 56; // the number of bits
printf("Leading 0's = %lld\n", i);
I'd be curious to see how this was efficiency wise. Tested it with several values though and it seems to work.
Based on: http://www.hackersdelight.org/HDcode/nlz.c.txt
template<typename T> int clz(T v) {int n=sizeof(T)*8;int c=n;while (n){n>>=1;if (v>>n) c-=n,v>>=n;}return c-v;}
If you'd like a version that allows you to keep your lunch down, here you go:
int clz(uint64_t v) {
int n=64,c=64;
while (n) {
n>>=1;
if (v>>n) c-=n,v>>=n;
}
return c-v;
}
As you'll see, you can save cycles on this by careful analysis of the assembler, but the strategy here is not a terrible one. The while loop will operate Lg[64]=6 times; each time it will convert the problem into one of counting the number of leading bits on an integer of half the size.
The if statement inside the while loop asks the question: "can i represent this integer in half as many bits", or analogously, "if i cut this in half, have i lost it?". After the if() payload completes, our number will always be in the lowest n bits.
At the final stage, v is either 0 or 1, and this completes the calculation correctly.
If you are dealing with unsigned integers, you could do this:
#include <math.h>
int numunset(uint64_t number)
{
int nbits = sizeof(uint64_t)*8;
if(number == 0)
return nbits;
int first_set = floor(log2(number));
return nbits - first_set - 1;
}
I don't know how it will compare in performance to the loop and count methods that have already been offered because log2() could be expensive.
Edit:
This could cause some problems with high-valued integers since the log2() function is casting to double and some numerical issues may arise. You could use the log2l() function that works with long double. A better solution would be to use an integer log2() function as in this question.
// clear all bits except the lowest set bit
x &= -x;
// if x==0, add 0, otherwise add x - 1.
// This sets all bits below the one set above to 1.
x+= (-(x==0))&(x - 1);
return 64 - count_bits_set(x);
Where count_bits_set is the fastest version of counting bits you can find. See https://graphics.stanford.edu/~seander/bithacks.html#CountBitsSetParallel for various bit counting techniques.
I'm not sure I understood the problem correctly. I think you have a 64bit value and want to find the number of leading zeros in it.
One way would be to find the most significant bit and simply subtract its position from 63 (assuming lowest bit is bit 0). You can find out the most significant bit by testing whether a bit is set from within a loop over all 64 bits.
Another way might be to use the (non-standard) __builtin_clz in gcc.
I agree with the binary search idea. However two points are important here:
The range of valid answers to your question is from 0 to 64 inclusive. In other words - there may be 65 different answers to the question. I think (almost sure) all who posted the "binary search" solution missed this point, hence they'll get wrong answer for either zero or a number with the MSB bit on.
If speed is critical - you may want to avoid the loop. There's an elegant way to achieve this using templates.
The following template stuff finds the MSB correctly of any unsigned type variable.
// helper
template <int bits, typename T>
bool IsBitReached(T x)
{
const T cmp = T(1) << (bits ? (bits-1) : 0);
return (x >= cmp);
}
template <int bits, typename T>
int FindMsbInternal(T x)
{
if (!bits)
return 0;
int ret;
if (IsBitReached<bits>(x))
{
ret = bits;
x >>= bits;
} else
ret = 0;
return ret + FindMsbInternal<bits/2, T>(x);
}
// Main routine
template <typename T>
int FindMsb(T x)
{
const int bits = sizeof(T) * 8;
if (IsBitReached<bits>(x))
return bits;
return FindMsbInternal<bits/2>(x);
}
Here you go, pretty trivial to update as you need for other sizes...
int bits_left(unsigned long long value)
{
static unsigned long long mask = 0x8000000000000000;
int c = 64;
// doh
if (value == 0)
return c;
// check byte by byte to see what has been set
if (value & 0xFF00000000000000)
c = 0;
else if (value & 0x00FF000000000000)
c = 8;
else if (value & 0x0000FF0000000000)
c = 16;
else if (value & 0x000000FF00000000)
c = 24;
else if (value & 0x00000000FF000000)
c = 32;
else if (value & 0x0000000000FF0000)
c = 40;
else if (value & 0x000000000000FF00)
c = 48;
else if (value & 0x00000000000000FF)
c = 56;
// skip
value <<= c;
while(!(value & mask))
{
value <<= 1;
c++;
}
return c;
}
Same idea as user470379's, but counting down ...
Assume all 64 bits are unset. While value is larger than 0 keep shifting the value right and decrementing number of unset bits:
/* untested */
int countunsetbits(uint64_t val) {
int x = 64;
while (val) { x--; val >>= 1; }
return x;
}
Try
int countBits(int value)
{
int result = sizeof(value) * CHAR_BITS; // should be 64
while(value != 0)
{
--result;
value = value >> 1; // Remove bottom bits until all 1 are gone.
}
return result;
}
Use log base 2 to get you the most significant digit which is 1.
log(2) = 1, meaning 0b10 -> 1
log(4) = 2, 5-7 => 2.xx, or 0b100 -> 2
log(8) = 3, 9-15 => 3.xx, 0b1000 -> 3
log(16) = 4 you get the idea
and so on...
The numbers in between become fractions of the log result. So typecasting the value to an int gives you the most significant digit.
Once you get this number, say b, the simple 64 - n will be the answer.
function get_pos_msd(int n){
return int(log2(n))
}
last_zero = 64 - get_pos_msd(n)
I need a little help with bitmap operations in C#
I want to take a UInt16, isolate an arbitrary number of bits, and set them using another UInt16 value.
Example:
10101010 -- Original Value
00001100 -- Mask - Isolates bits 2 and 3
Input Output
00000000 -- 10100010
00000100 -- 10100110
00001000 -- 10101010
00001100 -- 10101110
^^
It seems like you want:
(orig & ~mask) | (input & mask)
The first half zeroes the bits of orig which are in mask. Then you do a bitwise OR against the bits from input that are in mask.
newValue = (originalValue & ~mask) | (inputValue & mask);
originalValue -> 10101010
inputValue -> 00001000
mask -> 00001100
~mask -> 11110011
(originalValue & ~mask)
10101010
& 11110011
----------
10100010
^^
Cleared isolated bits from the original value
(inputValue & mask)
00001000
& 00001100
----------
00001000
newValue =
10100010
| 00001000
----------
10101010
Something like this?
static ushort Transform(ushort value){
return (ushort)(value & 0x0C/*00001100*/ | 0xA2/*10100010*/);
}
This will convert all your sample inputs to your sample outputs. To be more general, you'd want something like this:
static ushort Transform(ushort input, ushort mask, ushort bitsToSet){
return (ushort)(input & mask | bitsToSet & ~mask);
}
And you would call this with:
Transform(input, 0x0C, 0xA2);
For the equivalent behavior of the first function.
A number of the terser solutions here look plausible, especially JS Bangs', but don't forget that you also have a handy BitArray collection to use in the System.Collections namespace: http://msdn.microsoft.com/en-us/library/system.collections.bitarray.aspx
If you want to do bitwise manipulations, I have written a very versatile method to copy any number of bits from one byte (source byte) to another byte (target byte). The bits can be put to another starting bit in the target byte.
In this example, I want to copy 3 bits (bitCount=3) from bit #4 (sourceStartBit) to bit #3 (destinationStartBit). Please note that the numbering of bits starts with "0" and that in my method, the numbering starts with the most significant bit = 0 (reading from left to right).
byte source = 0b10001110;
byte destination = 0b10110001;
byte result = CopyByteIntoByte(source, destination, 4, 1, 3);
Console.WriteLine("The binary result: " + Convert.ToString(result, toBase: 2));
//The binary result: 11110001
byte CopyByteIntoByte(byte sourceByte, byte destinationByte, int sourceStartBit, int destStartBit, int bitCount)
{
int[] mask = { 0, 1, 3, 7, 15, 31, 63, 127, 255 };
byte sourceMask = (byte)(mask[bitCount] << (8 - sourceStartBit - bitCount));
byte destinationMask = (byte)(~(mask[bitCount] << (8-destStartBit - bitCount)));
byte destinationToCopy = (byte)(destinationByte & destinationMask);
int diff = destStartBit - sourceStartBit;
byte sourceToCopy;
if(diff > 0)
{
sourceToCopy = (byte)((sourceByte & sourceMask) >> (diff));
}
else
{
sourceToCopy = (byte)((sourceByte & sourceMask) << (diff * (-1)));
}
return (byte)(sourceToCopy | destinationToCopy);
}