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Here is one way I came up with -- though will work only for unsigned numbers ..

#include <stdio.h>

void main()
{
unsigned int i = 0xABCD;
unsigned int shift = 0;
unsigned int temp = 0;
unsigned reversed = 0;
//Check all nibbles.
for( int nCount = 0; nCount < sizeof(i) * 2; nCount++ )
{
temp = i >> shift;
//just keep the nibble.
temp = temp & (0 | 0xF);
//temp is now from 0 - F.
reversed = reversed | ((0xF-temp) << shift);
shift += 4;
}
printf("Original: %x\n", i);
printf("Reversed: %x\n", reversed );
}

can anybody explain in detail whats exactly going on here ?

how about this simle approach..

// v is value & r is reverse value

r = v
for (v >>= 1; v; v >>= 1)
{
r <<= 1;
r |= v & 1;
}
r <<= s; // shift when v's highest bits are zero

uint8_t a,b; // reverse a to b
b = 0;
for(int i =0;i<8;i++)
b |= ((a>>1)&0x01) << (7-i);

int res = 0;
	int rem;
	int n = 0xd;
	while(n)
	{
		rem = n%2;
		n >>= 1;
		res = (res<<1) | rem;

}

Will this work? (only for unsigned numbers)

int result = 0;
int a = 0;
while (n)
{
  a = n&1;
  result = result | a;
  result<<1;
  n>>1;
}

Use this:-
Reverse_Number = Number xor 0xFFFFFFFF;

wouldn't this swap the individual bits?
we are supposed to "reverse" the bits, taking into consideration all 32 bits at a time, and not swap each and every bit..

unsigned int reverse_binary(int n)
{
    unsigned int temp = n, result = 0;

    while (temp != 0) {
        if(temp & 1)
            result = result | 1;
        
        result <<= 1;
        temp >>= 1;
    }
    result >>= 1;
    return result;
}

public class ReverseBit {
	public static void main(String args[]) {
		int no = 10;
		int res = 0;
		int bit;
		// print before reverse
		printBits(no);
		// go thru the int from bit 1 to 32
		for (int i = 0; i < 32; i++) {
			// extract bit i
			bit = (no & (1 << i)) >>> i;
			// push into result with shifting
			res = (res << 1) | bit ;
		}
		// print bits after reversing
		printBits(res);
	}

	public static void printBits(int no) {
		System.out.println();
		for (int i = 31; i >= 0; i--)
			System.out.print((no & (1 << i)) >>> i);
	}
}

unsigned char revbits[16] = {0, 8, 4, 0xc, 2, 0xa, 6, 0xe, 1, 9, 5, 0xd, 3, 0xb, 7, 0xf};

int reversebits(int v)
{
int i = 0;
int cur = 0;
while(v)
{
i = (i << 4) | revbits[v & 0xF];
v >>= 4;
cur += 4;
}

return i << (32-cur);
}

Another solution using bitwise xor operation with 16 iterations is,

int reverse (int value)
{
unsigned int tmp = (unsigned int) value;
unsigned int bitcnt = 0;

while(bitcnt < 16)
{
/* Only change bits if the bit in lower 16 bits and
* corresponding bit in higher 16 bits are different
*/
if( ((tmp>>(31-bitcnt)) & 0x1) ^ ((tmp>>bitcnt) & 0x1) )
{
/* Bits differ so flip them */
tmp ^= 1<<(31-bitcnt) | 1<<bitcnt;
}
bitcnt++;
}
return((int)tmp);
}

I have tried the following program:


#include <stdlib.h>
#include <stdio.h>

void reverse (int );

int main()
{
int value;

printf ("\nEnter the Integer to be reversed: ");
scanf ("%d", &value);

reverse (value);
}

void reverse (int value)
{
int mask = 1 << 31;
int i;

for (i=0; i<=31; i++)
{
putchar (value & mask ? '1' : '0');
value <<= 1;
}
printf ("\nHope this helps!");
}

I have tried the following program:


#include <stdlib.h>
#include <stdio.h>

void reverse (int );

int main()
{
int value;

printf ("\nEnter the Integer to be reversed: ");
scanf ("%d", &value);

reverse (value);
}

void reverse (int value)
{
int mask = 1 << 31;
int i;

for (i=0; i<=31; i++)
{
putchar (value & mask ? '1' : '0');
value <<= 1;
}
printf ("\nHope this helps!");
}

I have tried the following program:

#include <stdlib.h>
#include <stdio.h>

void reverse (int );

int main()
{
   int value;

   printf ("\nEnter the Integer to be reversed: ");
   scanf  ("%d", &value);
   
   reverse (value);
}

void reverse (int value)
{
   int mask = 1 << 31;
   int i;
  
   for (i=0; i<=31; i++)
   {
    putchar (value & mask ? '1' : '0');
    value <<= 1;
   }
    printf ("\nHope this helps!");
}