How to extract specific bits from a number in C?

Question

I need to extract specific part (no of bits) of a short data type in C.

For Example I have a binary of 52504 as 11001101000 11000 and I want First 6 ( FROM LSB --> MSB i.e 011000 decimal 24) bits and rest of 10 bits ( 11001101000 decimal 820).

Similarly I want this function to be too generalized to extract specific no of bits given "start" and "end" (i.e chunks of bits equivalent with some decimal value).

I checked other posts, but those were not helpful, as given functions are not too much generalized.

I need something that can work for short data type of C.

Edit

I am having the short array of size 2048 bytes. Where each Pixel is of 10 bits. So my 16 bit consisting each byte occupying some time 2 pixels data, sometimes 3 pixels data.

Like

( PIXEL : 0,1 ) 10 BITS + 6 BITS

then ( PIXEL : 1,2,3 ) 4 BITS ( 1st pixels remaining bits ) + 10 BITS + 2 BITS.

and so on ..this pattern continues ... So, all I want to extract each pixel and make an entire array of having each pixels to be occupied wholy in on WHOLE BYTE ( of 16 bits ) like.. 1 byte should contain 1 DATA PIXEL, the other BYTE should contain other PIXEL value in whole 16 bits and so on so forth.

Conqueror

just mask what you want i.e. num & 0000000000111111 or if you want the other end (num & 1111111111000000) >> 6 – Frowsty

@hexa: you're probably right, but the question is a constructive question. What to do...? What to do...? – Chavannes

Can you make the question a bit more specific please? First you say you want the two numbers from splitting into two parts, then you say you want to specify a start and an end, does this mean you want 3 numbers out? Top, middle, and bottom? – Height

Actually i am processing the huge array of 2048 size of short carrying specific values of some pixels of an image taken from camera. Look at the edit – Conqueror

C types don't guarantee the size. short may have 18 bits or 32 bits or any value that conforms to the C standard. You need to specify a specific size, or use sized ints such as uint8_t... – Outfield

Answer 1

There are two building blocks that you need to know to build this yourself:

• Getting N least significant bits requires constructing a bit mask with N ones at the end. You do it like this: ((1 << N)-1). 1 << N is 2 ^ N: it has a single 1 at the N+1st position, and all zeros after it. Subtracting one gives you the mask that you need.
• Dropping M least significant bits is a simple shift to the right: k >> M

Now your algorithm for cutting out from M to N becomes a two-step process: you shift the original value M bits to the right, and then perform a bit-wise AND with the mask of N-M ones.

#define LAST(k,n) ((k) & ((1<<(n))-1))
#define MID(k,m,n) LAST((k)>>(m),((n)-(m)))

int main() {
    int a = 0xdeadbeef;
    printf("%x\n",  MID(a,4,16));
    return 0;
}

This fragment cuts out bits from 4, inclusive, to 16, exclusive, and prints bee when you run it. Bits are numbered from zero.

Pacien

Answer 2

unsigned short extract(unsigned short value, int begin, int end)
{
    unsigned short mask = (1 << (end - begin)) - 1;
    return (value >> begin) & mask;
}

Note that [begin, end) is a half open interval.

Evaporimeter

Answer 3

It can be done like this:

mask = ~(~0 << (end - start + 1));
value = (n >> start) & mask;

where n is the original integer and value is the extracted bits.

The mask is constructed like this:

1. ~0 = 1111 1111 1111 1111 1111 1111 1111 1111
2. ~0 << (end - start + 1) = 1111 1111 1111 1111 1100 0000 0000 0000
   // assuming we are extracting 14 bits, the +1 is added for inclusive selection
   // ensure that end >= start
3. ~(~0 << (end - start + 1)) = 0000 0000 0000 0000 0011 1111 1111 1111

Now n is shifted right by start bits to align the desired bits to the left. Then a bitwise AND gives the result.

Margo

returns wrong values.. For example..i have no 150 and its binary is 0000000010010110 ..I would need from start ( 2--5) bits..and they summs up with 22 . But this is not working at all. – Conqueror

@Conqueror 22 is the rightmost 5 bits... did you do it correctly? – Margo

yes of course either you pick First 5 bits or start from 2nd bit till 6th bit as i said earlier..it counts up to 22. So, I gave start = 1 and end = 5 or you can give it to end = 6...It MUST be 22. But its not returning 22, instead it returns 11 on start = 1 and end = 5. – Conqueror

I don,t understand.. All I need specific parts of the binar bits.. I think I have edited the post to make it more clear..:-) – Conqueror

@Conqueror But how come the bits 2 to 5 becomes 22? I don't understand that too. Bits 2 to 5 are 0101, right? 0000000010[0101]10 <- is the selection correct? – Margo

Nops..I am considering the first bit as the most LSB rightest bit that is '0' and then the second bit is '1' , the third would be again '1' and 0. So.. in 10110..( I need 1011 that is starting in original sequence given above of 150, starts from 2nd bit to 5th bit). – Conqueror

let us continue this discussion in chat – Margo

its giving wrong output, if i have 15 num and 1 to 2 bits (means 2nd and 3rd bit) i have to take out then it is giving 3 (0011) as answer, but the answer should come 6 (0110) – Chalkboard

Answer 4

//To get value from specific position 'pos' to 'pos+offset' in number 'value'

#define bitGet(value, offset, pos) (((1ull << offset) - 1) & (value >> (pos - 1)))

//Set value 'newval' from position 'pos' to 'pos+offset' in number 'value'

#define bitSet(value, offset, pos, newval)  \
(~(((1ull << offset) - 1) << (pos - 1)) & value) | ((((1ull << offset) - 1) & newval) << (pos - 1))

Cienfuegos

Answer 5

Although its a very old question, I would like to add a different solution. Using macros,

/* Here, startBit : start bit position(count from LSB) endBit : end bit position(count from LSB) .NOTE: endBit>startBit number : the number from which to extract bits maxLength:the total bit size of number. */ `

#include <stdio.h>
#define getnbits(startBit,endBit,number,maxLength) \
  ( number &  ( (~0U >> (maxLength-endBit)) & (~0U << startBit) )  ) 

int main()
{
    unsigned int num=255;
    unsigned int start=1,end=5,size=sizeof(num)*8;

    printf("Inputs : %d %d %d %d \n ",start,end,num,size);
    printf("Input number : %d\n",num);

    if(end>start)
    {
        int result = getnbits(start,end,num,size-1);
        printf("Output : %u\n\n",result);
    }
    else
        printf("Error : EndBit is smaller than starBit!\n\n");

    return 0;
}

`

Output : Inputs : 1 5 255 32
Input number : 255
Output : 62

Here, 255 = 11111111 and 62 = 00111110

Pupillary

Answer 6

// This is the main project file for VC++ application project 
// generated using an Application Wizard.

#include "stdafx.h"

#using <mscorlib.dll>

using namespace System;


void fun2(int *parr)
{
    printf(" size of array is %d\n",sizeof(parr));
}
void fun1(void)
{
    int arr[100];
    printf(" size of array is %d\n",sizeof(arr));
    fun2(arr);
}

int extractBit(int byte, int pos) 
{
    if( !((pos >= 0) && (pos < 16)) )
    {
        return 0;
    }
    return ( ( byte & (1<<pos) ) >> pos);
}
int extractBitRange(int byte, int startingPos, int offset) 
{


   if(  !(((startingPos + offset) >= 0) && ( (startingPos + offset) < 16)) )
   {
        return 0;
   }
   return ( byte >> startingPos ) & ~(0xff << (offset + 1));
}

int _tmain()
{
    // TODO: Please replace the sample code below with your own.

    int value;
    signed int res,bit;
    signed int stPos, len;
    value = 0x1155;
    printf("%x\n",value);
    //Console::WriteLine("Hello World");
    //fun1();
    for(bit=15;bit>=0;bit--)
    {
        res =extractBit(value,bit);
        printf("%d",res);
    }
    stPos = 4;
    len = 5;
    res = extractBitRange(value, stPos, len);
    printf("\n%x",res);

    return 0;
}

Seamanlike

Answer 7

void  f(short int last, short int first, short int myNr){
      //construct mask for last bits
      short int mask=0;
      for(int i=0;i<last;i++)
       { mask+=1;
        mask<<1;}
      short int aux= myNr;
      aux=aux&mask; // only last bits are left
      //construct mask for first bits
      mask=0;
      for(int i=0;i<first;i++)
       { mask+=0x8000h;
        mask>>1;} 
      aux=myNr;  
      aux&=mask;
      aux>>last; // only first bits are left and shifted
}

you can add parameters to get the values out or something

Sunbonnet

Answer 8

unsigned int extract_n2mbits(unsigned int x, int n, int m)
{
unsigned int mask, tmp;
if (n < m) {
    n = n + m;
    m = n - m;
    n = n - m;
}
mask = 1 << (n - m + 1);
tmp = m;
while (tmp > 1) {
    mask = mask << 1 | 1 << (n - m + 1);
    tmp = tmp - 1;
}
return ((x & mask) >> (n - m + 1));
}

Mindful

Answer 9

I know this is a very old question but obviously still relevent. Here is my contribution using the ((1 << N)-1) trick for getting the desrired bitmask. It wasn't asked here, but I'll offer up both ways to extract and store the desired bits. Let's say that someValue holds a 2-bit field called foo at bits 0-1 and a 3-bit field called bar at bits 3-5. Let's further assume you've got a code generator to hardcode the offsets into the setters and getters to simplify the interface because of course you do.

uint8_t someValue = 0;

inline uint8_t getFoo() const {
  //start bit = 0, size in bits = 2, end = 2
  uint8_t mask = (1 << (2 - 0)) - 1; //end-start-1
  return (someValue >> 0) & mask; //shift back by start
}

inline void setFoo(uint8_t foo) {
  someValue &= ~(((1 << (2 - 0)) - 1) << 0); //end-start-1, shift by start
  someValue |= foo << 0; //shift by start
}

inline uint8_t getBar() const {
  //start bit = 3, size in bits = 3, end = 6
  uint8_t mask = (1 << (6 - 3)) - 1; //end-start-1
  return (someValue >> 3) & mask; //shift back by start
}

inline void setBar(uint8_t bar) {
  someValue &= ~(((1 << (6 - 3)) - 1) << 3); //end-start-1, shift by start
  someValue |= bar << 3; //shift by start bit
}

Sausa