Bit Usage

You can add/subtract a mask...

For example:

Code:

#include <iostream>

int ToggleBit(int number, int bit_num)
{
	
	int mask = 1;
	
	for(int loop = 0; loop < bit_num; loop++)
		mask *= 2;
	
	if((number & mask) != 0)
		number -= mask;
	
	else number += mask;
	
	return number;
	
}

int main()
{
	
	std::cout << std::hex;
	std::cout << ToggleBit(0x20, 3);
	
	std::cin.get();
	return 0;
	
}

And my output is

28

It toggled the 3rd bit.

I put together a little function which may help you understand it better.

Code:

/*
 * This function will set or clear a particular bit.
 * Parameters:
 *		bit - the bit to change.  1 is least significant.
 *		someInt - pointer to integer to operate on.
 *		set - tells whether to set or clear the bit
 */
void ChangeBitValue(int bit, int *someInt, int set)
{
	int mask = 1;

	mask = mask << (bit - 1);

	if(set)
		*someInt = *someInt | mask;
	else
		*someInt = *someInt & ~mask;
}

lol.

As, you can see, there are many different ways to go about bit manipulation.

Thanx.
Is there any way i can get the system to store an integer in 3 bytes.
And how do i do consecutive reads on a two dimensional array.
eg:- If i had an array[8][9] how can i read it 6 values at a time. That is i want to read the whole array, by reading 6 values of it at a time.
-Cheers.

Is there any way i can get the system to store an integer in 3 bytes.

I'm sure you could do it manually with a union or an array of 3 chars (with bitwise operations, of course).

And how do i do consecutive reads on a two dimensional array.

It's all in how you manage the destination array and the loop that processes it. That's all I can tell you short of doing the problem for you.

And no I'm not going to do the problem for you.

>Is there any way i can get the system to store an integer in 3 bytes.
Yes, but you would still probably be using at least 4 bytes of storage and why would you want to do this anyway?

>That is i want to read the whole array, by reading 6 values of it at a time.
Hmm, something like this perhaps?

Code:

#include <stdio.h>

int main ( void )
{
  int a[12];
  int i = 0;
  int n = 0;

  while ( i < 12 ) {
    /* Eew */
    i += n = scanf ( "%d %d %d %d %d %d", &a[i], &a[i + 1], &a[i + 2], &a[i + 3], &a[i + 4], &a[i + 5] );

    if ( n != 6 )
      break;
  }

  while ( --i >= 0 )
    printf ( "%d\n", a[i] );

  return 0;
}