Hi guys,
This is probably pretty basic... but I don't seem to get it:
How does
(2 & 1) = 0
(3 & 1) = 1
(4 & 1) = 0
etc..
This pattern above seems to help find even numbers
or
(0 | 1) = 1
(1 | 1) = 1
(2 | 1) = 3
(3 | 1) = 4
(4 | 1) = 5
(5 | 1) = 5
I know how boolean algebra works between bits. But I don't understand how Boolean algebra works with integers (in C# at the least).
thanks in advance.
It works the same way in C# as it does in binary.
2 | 1 = 3 and 4 | 1 = 5.
To understand this, you need to think about the binary representation of 1,2,3,4,and 5:
010 | 001 = 011 and 100 | 001 = 101.
Similarly:
010 & 001 = 000 and 011 & 001 = 001
It is doing bitwise operations on the integer. That it is doing a logical or/and of each bit in the first integer with the corresponding bit in the other integer. It then returns the result of all of these operations. For example, 4 = 0100 and 1 = 0001, a logical and of these would and bit the bits in order and get 0000 (since 0&0 = 0, 1&0 = 0, 0&0 = 0, and 0&1 = 0). For or, you would get 0101 (since 0|0 = 0, 1|0 = 1, 0|0 = 0, and 0|1 = 1). The trick is that these are bitwise operations, not logical operations which operate only on boolean values in C#.
The key is that the CPU is doing 32 boolean operations in parallel, one for each bit position of the input integers (assuming 32 bit integers, of course).
You are getting the first result because you are performing a boolean and between the bit strings of the two numbers:
2 & 1 => 010 & 001 = 000 = 0
3 & 1 => 011 & 001 = 001 = 1
4 & 1 => 100 & 001 = 000 = 0
5 & 1 => 101 & 001 = 001 = 1
In effect, you are testing whether the '1' bit is set, which will only be true for odd numbers.
When performing or operations:
0 | 1 => 000 | 001 = 001 = 1
1 | 1 => 001 | 001 = 001 = 1
2 | 1 => 010 | 001 = 011 = 3
3 | 1 => 011 | 001 = 011 = 3
4 | 1 => 100 | 001 = 101 = 5
5 | 1 => 101 | 001 = 101 = 5
Since in this case the effect or the or is to always set the 1 bit, even numbers will be incremented by one to their nearest greater odd number.