Thread: bitwise operators?

Evil,

Are you using that book that introduces bitwise operations in the 3rd chapter? (Horton or Schiltd???) I've forgotton what book that is, but that's nuts!!!! (It is something you will need to know eventually, but most C/C++ programs don't have any bitwise operations.)

The only reason this is hard is because binary is not built-into cin or cout.

The secret to binary/bitwise operations is: Hexadecimal.

Hex is built into cin and cout.

Hex is easier than decimal because, with a little practice you can learn to convert between hex and binary (any size numbers) in your head! It's one more thing to learn, but you only have to memorize the 16 conversions from 0 to F... and you already know 0 and 1.

In my "hardware world", we work with individual bits every day and we never mix binary and decimal! I might say "I'm reading FE", I would not say "I'm reading 254".

Binary -> Hex
0000 -> 0
0001 -> 1
0010 -> 2
0011 -> 3
0100 -> 4
0101 -> 5
0110 -> 6
0111 -> 7
1000 -> 8
1001 -> 9
1010 -> A
1011 -> B
1100 -> C
1101 -> D
1110 -> E
1111 -> F

BTW - The calculator built-into Windows can do conversions between binary, octal, decimal, and hex, if you set the view to Scientific.


Doug.

I know this is probably a bump, after such a long time since originally being posted. I thought I posted a reply to this and posted it, I don't know what happened.

Doug I understand how the book was doing this in binary now. The hexadecimal things seems easy enough to understand (Even if there is no source anywhere to learn it) so maybe I will learn it some other day.

Thank you everyone so much for your help!

An yes I know I am being a total lazy ass, I just cannot help it!

Maybe you could see it that way:

decimal: 5432

is like
(5 * (10^3)) + (4 * (10^2)) + (3 * (10^1)) + (2 * (10^0))
(5 * 1000) + (4 * 100) + (3 * 10) + (2 * 1)
5000 + 400 + 30 + 2
= 5432

so you just take the digits positions and add them together by their digits, such
as 5, being at the position 10^3, where the 10 represent the base, decimal, and 3, the
thousands multiplication to 5...

In all other bases, it works the same way, only the "10"s change for 16, 8, 2... depending on the current base.

So for binary (base 2),
1011
is like
(1 * (2^3)) + (0 * (2^2)) + (1 * (2^1)) + (1 * (2^0))
(1 * 8) + (0 * 4) + (1 * 2) + (1 * 1)
8 + 2 + 1
= 11

And for hexadecimal, (base 16) 0FA5
is like
(0 * (16^3)) + (F * (16^2)) + (A * (16^1)) + (5 * (16^0))
(0 * 4096) + (F * 256) + (A * 16) + (5 * 1)
(0 * 4096) + (15 * 256) + (10 * 16) + (5 * 1)
3840 + 160 + 5
= 4005

You have to know that 0 1 2 3 4 5 6 7 8 9 A B C D E F
Is equivalent in decimal to 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

You should be able to get something out of this...