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boolean isBitPalindrome(int num) {
		int store = num;
		int reversedNum = 0;
		while ( num != 0) {
			if ( (num & 1) == 1) {
				reversedNum += 1;
			}
			num = num >> 1;
			if ( num != 0)
				reversedNum = reversedNum << 1;
		}
		return (store == reversedNum);
	}

What questions were you asked in Multithreading?

lets take an example
0xabcddcba
Byte0 is Reverse of Byte3 (ba .. ab)
Byte1 is Reverse of Byte2 (dc ..cd)

Logic is simple, just compare B3 with reverse of B0 && B1 with reverse of B2. if both evaulates to TRUE, given bit pattern of an integer is a palindrome.

I assume there exists a lookup table which already stores the reverse of numbers from 0-256.

if((lookupTable[n&0xff]) == (n>>24 & 0xff)
&& (lookupTable[n>>8 & 0xff]) == (n>>16 & 0xff)
)
{
printf("Bit pattern of <%d> is a palindrome\n", n);
}

// copy the bit pattern backwards & xor with original.
// if result is 0 then its palindrome,return true
// else false
bool isPalindrome( int x )
{
size_t t = sizeof(int)*CHAR_BIT; // num of bits in the int
int x_rev = 0; // to store the reverse pattern
const int mask = 1;

for ( int i = 0; i < t; ++i ) {

// the bit at position t-i must
// be the bit at position i for a
// palindrome! 0 bits are already there
// in x_rev. Only 1s need to be
// determined
if ( x & (mask << i) ) {
x_rev |= mask << t - i;
}
}
return !(x^x_rev);
}

bool Ispalin(int b1){
	int b2;
	while(b1){
		b2 + = (b1 & 1);
		B1>>1;
		B2<<1;
	}
return ((b1==b2)? 1:0)

}

int check_palindrome(unsigned int n)
{
  unsigned int t = 0;

  for(unsigned int bit = 1; bit && bit <= n; bit <<= 1)
    t = (t << 1) | !!(n & bit);

  return t == n;
}