Bitwise AND on 32-bit Integer

问题:

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

Related:

Most common C# bitwise operations.

回答1:

With the & operator

回答2:

Use the & operator.

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

From MSDN.

回答3:

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

回答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 */ }

回答5:

use & operator (not &&)

回答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

回答7:

The & operator

回答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;