How to Manipulate Bits in C and C++ by@botman1001

# How to Manipulate Bits in C and C++

### @botman1001Abhishek Singh Thakur

All data in computer is represented in binary i.e. in 0 or 1. Computers or machines do not understand our languages, they understand bits. Generally programmer do not care about operations at the bit level. But sometimes a programmer has to dive in a deeper level and work on bits.

# Bits representation

In programming, an n bit integer is stored as a binary number that
consists of n bits. So a 32-bit integer consists of 32 bits and 64 bit
integer consists of 64 bits. In C++ programming language int data type
is 16-bit, 32-bit and 64-bit type. see here.

Here is the bit representation of 32 bit int number 10:
00000000000000000000000000001010

In C++,

``int``
is either signed or unsigned and so a bit representation is either signed or unsigned.

In a signed representation, first bit represents the sign of a number (0
for positive and 1 for negative), and remaining n-1 bits contains the
magnitude of the number.

There is a connection between a signed and an unsigned representation. A signed number

``-x ``
equals an unsigned number
``2^n – x``
.

``-x (signed) = 2^n - x (unsigned)``

``````int a = -10;
unsigned int b = a;
std::cout << a << "\n"; /* -10 */
std::cout << b << "\n"; /* 4294967286 */``````

In a signed representation, the next number after

``2^(n – 1) – 1``
is
``-2^n – 1``
, and in an unsigned representation, the next number after
``2^n – 1``
is
``0``
.

# Bit Operations

We can use & operator to check if a number is even or odd. If

``x & 1 = 0``
then
``x``
is even and if
``x & 1 = 1``
then
``x``
is odd. We can also say that,
``x``
is divisible by
``2^k``
exactly when
`` x & (2^k – 1) = 0.``

``x<<k ``
corresponds to multiplying
``x``
by
``2^k``
, and
``x>>k``
corresponds to dividing
``x``
by
``2^k``
rounded down to an integer.

# Common Bit Tasks

Binary representation of unsigned int:

``````void binary(unsigned int num)
{
for(int i = 256; i > 0; i = i/2) {
if(num & i)
std::cout << "1 ";
else
std::cout << "0 ";
}
std::cout << std::endl;
}``````

Setting Bit at position:

``````int set_bit(int num, int position)
{
int mask = 1 << position;
return num | mask;
}``````

Getting Bit at position:

``````bool get_bit(int num, int position)
{
bool bit = num & (1 << position);
return bit;
}``````

Clearing Bit at position:

``````int clear_bit(int num, int position)
{
int mask = 1 << position;
return num & ~mask;
}``````

# Representing Sets

Bits representation of an integer are 0-indexed and the index starts from right side i.e. least significant bit. So we can represent every subset of the set

``{0, 1, 2, ..., n-1}``
as an n bit integer and whose bits indicate which element belongs to the subset. If bit at index 3 is 1 and at index 4 is 0 in binary representation of a number, then 3 belongs to the subset and 4 does not belong to the subset.

For a 32-bit integer, the set is {0, 1, 2,…, 31} and the subset is {1, 3, 4, 8}. The binary representation of the set is:
00000000000000000000000100011010
and decimal representation is 2^8 + 2^4 + 2^3 + 2^1 = 282.

Code to form subset and add elements to it:

``````int add_elements_to_subset()
{
int subset = 0;
subset = subset | (1 << 1);
subset = subset | (1 << 3);
subset = subset | (1 << 4);
subset = subset | (1 << 8);
return subset;
}``````

Code to print elements of the subset:

``````void printing_subset(int subset)
{
for (int i = 0; i < 32; i++)
{
if (subset & (1 << i)) std::cout << i << " ";
}
}``````

The g++ compiler provides the following functions for counting bits:

``__builtin_clz(x)``
: the number of zeros at the beginning of the number

``__builtin_ctz(x)``
: the number of zeros at the end of the number

``__builtin_popcount(x)``
: the number of ones in the number

``__builtin_parity(x)``
: the parity (even or odd) of the number of ones

Originally posted at ProgrammerCave

``Reference: ``
``Competitive Programmer’s Handbook, by Antti Laaksonen``