Bitwise AND on 32-bit Integer

Question

How do you perform a bitwise AND operation on two 32-bit integers in C#?

Related:

Most common C# bitwise operations.

Answer 1

With the & operator

Answer 2

Use the & operator.

Binary & operators are predefined for the integral types[.] For integral types, & computes the bitwise AND of its operands.

From MSDN.

Answer 3

var x = 1 & 5;
//x will = 1

Answer 4

const uint 
  BIT_ONE = 1,
  BIT_TWO = 2,
  BIT_THREE = 4;

uint bits = BIT_ONE + BIT_TWO;

if((bits & BIT_TWO) == BIT_TWO){ /* do thing */ }

Answer 5

use & operator (not &&)

Answer 6

int a = 42;
int b = 21;
int result = a & b;

For a bit more info here's the first Google result:
http://weblogs.asp.net/alessandro/archive/2007/10/02/bitwise-operators-in-c-or-xor-and-amp-amp-not.aspx

Answer 7

The & operator

Answer 8

var result = (UInt32)1 & (UInt32)0x0000000F;

// result == (UInt32)1;
// result.GetType() : System.UInt32

If you try to cast the result to int, you probably get an overflow error starting from 0x80000000, Unchecked allows to avoid overflow errors that not so uncommon when working with the bit masks.

result = 0xFFFFFFFF;
Int32 result2;
unchecked
{
 result2 = (Int32)result;
}

// result2 == -1;