Archive for the ‘bit-operator programs’ Category

reversing the bits in an interger

August 31, 2007

#include<stdio.h>
#include<string.h>
#include<iostream>
using namespace std;
int main()
{
unsigned int num; // Reverse the bits in this number.
unsigned int temp = num; // temp will have the reversed bits of num.
cout<<“\n enter the num :”;
cin>>num;
int i;
for (i =(sizeof(num)*8-1); i; i–)
{
temp = temp | (num & 1);
temp <<= 1;
num >>= 1;
}
temp = temp | (num & 1);
cout<<temp;

cout<<endl<<endl;
system(“pause”);
return 0;
}
//2nd method

//lookup table method

const unsigned char ReverseLookupTable[] =
{
0x00, 0x80, 0x40, 0xC0, 0x20, 0xA0, 0x60, 0xE0, 0x10, 0x90, 0x50, 0xD0, 0x30, 0xB0, 0x70, 0xF0,
0x08, 0x88, 0x48, 0xC8, 0x28, 0xA8, 0x68, 0xE8, 0x18, 0x98, 0x58, 0xD8, 0x38, 0xB8, 0x78, 0xF8,
0x04, 0x84, 0x44, 0xC4, 0x24, 0xA4, 0x64, 0xE4, 0x14, 0x94, 0x54, 0xD4, 0x34, 0xB4, 0x74, 0xF4,
0x0C, 0x8C, 0x4C, 0xCC, 0x2C, 0xAC, 0x6C, 0xEC, 0x1C, 0x9C, 0x5C, 0xDC, 0x3C, 0xBC, 0x7C, 0xFC,
0x02, 0x82, 0x42, 0xC2, 0x22, 0xA2, 0x62, 0xE2, 0x12, 0x92, 0x52, 0xD2, 0x32, 0xB2, 0x72, 0xF2,
0x0A, 0x8A, 0x4A, 0xCA, 0x2A, 0xAA, 0x6A, 0xEA, 0x1A, 0x9A, 0x5A, 0xDA, 0x3A, 0xBA, 0x7A, 0xFA,
0x06, 0x86, 0x46, 0xC6, 0x26, 0xA6, 0x66, 0xE6, 0x16, 0x96, 0x56, 0xD6, 0x36, 0xB6, 0x76, 0xF6,
0x0E, 0x8E, 0x4E, 0xCE, 0x2E, 0xAE, 0x6E, 0xEE, 0x1E, 0x9E, 0x5E, 0xDE, 0x3E, 0xBE, 0x7E, 0xFE,
0x01, 0x81, 0x41, 0xC1, 0x21, 0xA1, 0x61, 0xE1, 0x11, 0x91, 0x51, 0xD1, 0x31, 0xB1, 0x71, 0xF1,
0x09, 0x89, 0x49, 0xC9, 0x29, 0xA9, 0x69, 0xE9, 0x19, 0x99, 0x59, 0xD9, 0x39, 0xB9, 0x79, 0xF9,
0x05, 0x85, 0x45, 0xC5, 0x25, 0xA5, 0x65, 0xE5, 0x15, 0x95, 0x55, 0xD5, 0x35, 0xB5, 0x75, 0xF5,
0x0D, 0x8D, 0x4D, 0xCD, 0x2D, 0xAD, 0x6D, 0xED, 0x1D, 0x9D, 0x5D, 0xDD, 0x3D, 0xBD, 0x7D, 0xFD,
0x03, 0x83, 0x43, 0xC3, 0x23, 0xA3, 0x63, 0xE3, 0x13, 0x93, 0x53, 0xD3, 0x33, 0xB3, 0x73, 0xF3,
0x0B, 0x8B, 0x4B, 0xCB, 0x2B, 0xAB, 0x6B, 0xEB, 0x1B, 0x9B, 0x5B, 0xDB, 0x3B, 0xBB, 0x7B, 0xFB,
0x07, 0x87, 0x47, 0xC7, 0x27, 0xA7, 0x67, 0xE7, 0x17, 0x97, 0x57, 0xD7, 0x37, 0xB7, 0x77, 0xF7,
0x0F, 0x8F, 0x4F, 0xCF, 0x2F, 0xAF, 0x6F, 0xEF, 0x1F, 0x9F, 0x5F, 0xDF, 0x3F, 0xBF, 0x7F, 0xFF
};
(table is taken from some sites):)
unsigned int num; // Reverse the bits in this number.
unsigned int temp; // Store the reversed result in this.

temp = (ReverseLookupTable[num & 0xff] << 24) |
(ReverseLookupTable[(num >> 8) & 0xff] << 16) |
(ReverseLookupTable[(num >> 16) & 0xff] << 8) |
(ReverseLookupTable[(num >> 24) & 0xff]);

temp is the reversed-bit number…

count no. of setbits in any given no.

August 28, 2007

#include <stdio.h>
#include<iostream>
using namespace std;

//1st method
int coutsetbits(int Num)
{
int count=0;
for( ; Num; Num>>=1)
{
if (Num & 1)
count++;
}
return count;
}

The operators >> and <<, dont change the operand at all. However, the operators >>= and <=< also change the operand after doing the shift operations.
//2nd method
int coutsetbits(int Num)
{
int count;
for( count =0; Num; count++)
{
Num = Num & Num -1;
}
return count;
}

//simplify it

{

while(num!=0)

{ count++;

num=num &(num-1) //clears last significant setbit

}

return count;

}

int main ()
{
int num;
cout<<“\n enter the num :”;
cin>>num;
int a;
a=coutsetbits(num);
cout<<a;
system(“pause”);
return 0;
}

//3rd method

write the num in binary form by dividing 2 each time till then it becam 0 n store it in any string/char-string..now count how many setbit r ?

int main ()
{
int num,i=0,a[50],count=0,count1=0;
cout<<“\n enter the num :”;
cin>>num;

while(num!=0)
{ a[i]=num%2; count++; i++; num=num/2; }

for(int j=count-1;j>=0;j–)
cout<<a[j];
for(int j=count-1;j>=0;j–)
{ if(a[j]==1)
count1++; }
cout<<“\n no. of set bit is :”<<count1;
cout<<endl<<endl;
system(“pause”);
return 0;
}

//4th method

This speeds up the computation. What it does is it keeps a table which hardcodes the number of bits set in each integer from 0 to 256.

For example


0 - 0 Bit(s) set.
1 - 1 Bit(s) set.
2 - 1 Bit(s) set.
3 - 2 Bit(s) set.
...

So here is the code…

const unsigned char LookupTable[] =

{ 0, 1, 1, 2, 1, 2, 2, 3, 1, 2, 2, 3, 2, 3, 3, 4,
1, 2, 2, 3, 2, 3, 3, 4, 2, 3, 3, 4, 3, 4, 4, 5,
1, 2, 2, 3, 2, 3, 3, 4, 2, 3, 3, 4, 3, 4, 4, 5,
2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6,
1, 2, 2, 3, 2, 3, 3, 4, 2, 3, 3, 4, 3, 4, 4, 5,
2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6,
2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6,
3, 4, 4, 5, 4, 5, 5, 6, 4, 5, 5, 6, 5, 6, 6, 7,
1, 2, 2, 3, 2, 3, 3, 4, 2, 3, 3, 4, 3, 4, 4, 5,
2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6,
2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6,
3, 4, 4, 5, 4, 5, 5, 6, 4, 5, 5, 6, 5, 6, 6, 7,
2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6,
3, 4, 4, 5, 4, 5, 5, 6, 4, 5, 5, 6, 5, 6, 6, 7,
3, 4, 4, 5, 4, 5, 5, 6, 4, 5, 5, 6, 5, 6, 6, 7,
4, 5, 5, 6, 5, 6, 6, 7, 5, 6, 6, 7, 6, 7, 7, 8};

//table is taken from a site

unsigned int num;
unsigned int ctr; // Number of bits set.
//0xff==(11111111)(in binary form)

//it'll check 1 byte(8 bit) at a time of any int so we extract 1 byte at a time by rightshift operator n find the corresponding value from lookuptable..

ctr = LookupTable[num & 0xff] +
LookupTable[(num >> 8) & 0xff] +
LookupTable[(num >> 16) & 0xff] +
LoopupTable[num >> 24];