Get a specific bit from byte
Solution 1:
Easy. Use a bitwise AND to compare your number with the value 2^bitNumber, which can be cheaply calculated by bit-shifting.
//your black magic
var bit = (b & (1 << bitNumber-1)) != 0;
EDIT: To add a little more detail because there are a lot of similar answers with no explanation:
A bitwise AND compares each number, bit-by-bit, using an AND join to produce a number that is the combination of bits where both the first bit and second bit in that place were set. Here's the logic matrix of AND logic in a "nibble" that shows the operation of a bitwise AND:
0101
& 0011
----
0001 //Only the last bit is set, because only the last bit of both summands were set
In your case, we compare the number you passed with a number that has only the bit you want to look for set. Let's say you're looking for the fourth bit:
11010010
& 00001000
--------
00000000 //== 0, so the bit is not set
11011010
& 00001000
--------
00001000 //!= 0, so the bit is set
Bit-shifting, to produce the number we want to compare against, is exactly what it sounds like: take the number, represented as a set of bits, and shift those bits left or right by a certain number of places. Because these are binary numbers and so each bit is one greater power-of-two than the one to its right, bit-shifting to the left is equivalent to doubling the number once for each place that is shifted, equivalent to multiplying the number by 2^x. In your example, looking for the fourth bit, we perform:
1 (2^0) << (4-1) == 8 (2^3)
00000001 << (4-1) == 00001000
Now you know how it's done, what's going on at the low level, and why it works.
Solution 2:
While it's good to read and understand Josh's answer, you'll probably be happier using the class Microsoft provided for this purpose: System.Collections.BitArray It's available in all versions of .NET Framework.
Solution 3:
This
public static bool GetBit(this byte b, int bitNumber) {
return (b & (1 << bitNumber)) != 0;
}
should do it, I think.