Reversing a binary number

navenmou · Aug 17, 2011

Hi all

I want to write a C program for reversing a binary number i.e like this 001101---->101100.....I am not getting idea on how to proceed....please give me some ideas for that....

Thanks in advance

ckshivaram · Aug 17, 2011

Re: Revrsing a binary number

check this

Write an Efficient C Program to Reverse Bits of a Number | GeeksforGeeks

Bit Twiddling Hacks

Code:

#include <stdio.h>  

int main ()  
{  

   int r0, myIn=0,myIn1=0;  

   printf("Please enter a number: ");  
   scanf("%d", &myIn);  

   do  
   {  
      r0 = myIn%2;  /* get remainder value*/ 

     myIn = (myIn-r0)/2; /* (initial value - remainder) /2 */ 
     printf("%d",r0);  
    }
       while(myIn>0);  
       return 0;

bigdogguru · Aug 17, 2011

Re: Revrsing a binary number

Here's a quick and efficient algorithm to reverse the ordering of a byte:

Code:

unsigned char rbyte = 0;
unsigned char byte = 0xAA; // 0xAA Represents Value to Be Reversed

int i;

for (i=0 ; i<4 ; i++) 
rbyte |= ((byte & (1 << i)) << (7-2*i)) | ((byte & (1 << 7-i)) >> (7-2*i));

It works well.

BigDog

alexan_e · Aug 17, 2011

Re: Revrsing a binary number

bigdogguru said:
Here's a quick and efficient algorithm to reverse the ordering of a byte:

Code:

unsigned char rbyte = 0; unsigned char byte = 0xAA; // 0xAA Represents Value to Be Reversed int i; for (i=0 ; i<4 ; i++) rbyte |= ((byte & (1 << i)) << (7-2*i)) | ((byte & (1 << 7-i)) >> (7-2*i));

It works well.

BigDog

Another option would be

Code C - [expand]
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unsigned char rbyte = 0;
unsigned char byte = 0xAA; // 0xAA Represents Value to Be Reversed
 
int i;
 
for (i=0 ; i<8 ; i++)
    if (byte & (1<<i)) rbyte |= (1<<(7-i));

I have tested it in AVRstudio and it executes at about 20% less cpu cycles

Alex

FvM · Aug 17, 2011

Re: Revrsing a binary number

If the bit reversal has to be carried out frequently in your code, a ROM table is the fastest solution, at least for 8 bit entities. For larger numbers, byte swap can be used in addition.

alexan_e · Aug 18, 2011

Re: Revrsing a binary number

This is an example of the lookup method suggested by FvM applied to 8-bit char and 16-bit integer, the lookup table is used for each nibble (4-bits)

Code C - [expand]
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unsigned char reverse_byte (unsigned char input_byte)
{
  const unsigned char lookup[16]={0,8,4,12,2,10,6,14,1,9,5,13,3,11,7,15}; // depending on the compiler use the proper keyword to make this a flash (ROM) variable
 
  return (lookup[input_byte & 15]<<4) | (lookup[input_byte>>4]);
}
 
 
 
unsigned int reverse_int (unsigned int input_int)
{
  const unsigned char lookup[16]={0,8,4,12,2,10,6,14,1,9,5,13,3,11,7,15}; // depending on the compiler use the proper keyword to make this a flash (ROM) variable
 
  return (lookup[input_int & 15]<<12) | (lookup[(input_int>>4) & 15]<<8) | (lookup[(input_int>>8) & 15]<<4) | (lookup[(input_int>>12)]);
}

Alex

saeednd · Jun 21, 2012

Revrsing a binary number Tested in Codevision 100% work,By Saeedvand.

void revise_data(char byte)
{
unsigned char hbyte,lbyte,d; // 0xAA Represents Value to Be Reversed
char revise,Final_revise_data;
int i;

revise=0;
lbyte=(byte)%(0x10);
hbyte=byte-lbyte;
for (i=0 ; i<4 ; i++)
revise |= ((lbyte & (1 << i)) << (7-2*i)) | ((lbyte & (1 << 7-i)) >> (7-2*i));
lbyte=revise;

revise=0;
for (i=0 ; i<4 ; i++)
revise |= ((hbyte & (1 << i)) << (7-2*i)) | ((hbyte & (1 << 7-i)) >> (7-2*i));

hbyte=revise;

Final_revise_data=hbyte+lbyte;

}

Reversing a binary number

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Code C - [expand]
1 2 3 4 5 6 7	unsigned char rbyte = 0; unsigned char byte = 0xAA; // 0xAA Represents Value to Be Reversed int i; for (i=0 ; i<8 ; i++) if (byte & (1<<i)) rbyte \|= (1<<(7-i));

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