Looking for the fastest way to divide by 2 - c++

Question

I've searched half the day and found some very interesting things about using fixed point data types and bit shifting in C++ to accomplish division operations while avoiding floating point math. However, I have only been able to understand a small fraction of it and I can't seem to get anything to work.
All I'm wanting to do is to take two integers, ad them up, and divide by two to get the average. I need to be able to do this very quickly though, since I'm interpolating camera pixel data on an Arduino and I also have other operations to do.
So I'm confused about shifting in general. Say the integer I want to divide by two is 27. Half of 27 is 13.5. But no matter what fixed point datatype I try, I can only get 13 as an output. For example:
uint8_t x = 27;
Serial.println( x >> 1 );
returns 13
There's got to be some simple way to do this, right?

Answer 1

Fixed point does give you a way to represent 13.5. The Wikipedia article on the Q number format is informative: https://en.wikipedia.org/wiki/Q_(number_format)
Think of it this way: You keep using integers, but instead of taking them at face value, divide them all implicitly by a power of 2 to obtain their semantic value.
So, if using an unsigned byte as your base type (values between 0 and 255, inclusive), you might implicitly divide by 2**3 (8). Now to represent 27, you need an integer set to 27*8=>216. To divide by two, you shift one to the right; now your integer is 108, which when divided by the implicit denominator of 8 gives 13.5, the value you're expecting.
You have to realize that fixed-point number systems (and floating point too, though it's less immediately evident) still have limits, of course; certain operations will overflow no matter what you do, and some operations cause a loss of precision. This is a normal consequence of working with limited-size types.

Answer 2

Say the integer I want to divide by two is 27. Half of 27 is 13.5. But
no matter what fixed point data type I try, I can only get 13 as an
output.
From wikipedia Fixed-Point Arithmetic:
The scaling factor is usually a power of 10 (for human convenience) or
a power of 2 (for computational efficiency).
You actually mentioned fixed point data type, and I think that is the best approach. But no matter what you tried? Perhaps we have different understandings of fixed-point-arithmetic.
while avoiding floating point math.
Another worth while goal, though reducing in value. Even in embedded systems, I seldom had to deal with a processor that did not have floating point parts. Floating point hardware has gotten reasonably good.
Any way, using fixed point avoids any need for floating point. Even for display purposes.
I think I need to proceed with a few examples.
Fixed point Example 1: Dollars and pennies
The unit of American money is based on the dollar. The Dollar is a fixed point data type.
So, if you have 27 dollars, how do you split it with your sibling?
One way (of several) that you all know is to convert 27 dollars into 2700 pennies. Dividing this value by 2 is trivial. Now you and your sibling can each get 1350 pennies. (i.e. the penny is a fixed point data type, that easily converts to/from dollars, and vice-vesa)
Note that this is completely integer arithmetic. Adding 2 integers, and dividing by 2 (any modern compiler will choose the fastest implementation.. either integer divide or perhaps right-shift-by-2) and on my desktop these 2 actions take less than a microsecond to complete.
You should waste no more time on measuring the relative performance of those two options (divide vs right-shift), you simply enable -O3 when your code tests correct. Your compiler should be able to choose correctly.
The choice of units in any problem is based on a scale factor that covers the range of values (in your problem) AND the understandable and quickly implemented conversion between units. And note that uint64_t can describe a large amount of cash, even in pennies. (challenge to the student.)
In General, about fixed point:
Given
uint8_t x = 27;
and the desire to divide by 2 evenly and quickly... can any scale factor be something that serves your needs? I say yes.
example 2 - 50 cent coins and a dollar
How about we try, for example, a simple scale factor of 2, i.e. the unit is a hu, or half unit. (analogous to the 50-cent-coin)
uint8_t x = 27 * 1/hu; (hu = 1/2)
This means that 54 hu represents 27 units. (ie, it takes 54 50-cent-coins to add up to 27 dollars)
The fixed point solution is to scale your integer values to achieve the arithmetic required. If you scale to even values, all your integers will divide evenly to the hu units.
example 3 - nickles and a dollar
Another possible scale might be 20, both decimal (for readability) and binary for performance. (note that there are 20 nickels in a dollar)
uint16 x = 27 * 1/tu; (tu = 1/20)
Now 540 represents a scaled 27. i.e. 540 nickles
All examples are fully integer, provide exact answers, and there is a trivial mechanism to convert the values for presentation to the user. i.e. which ever fixed point used, convert to analogue of pennies, and thus 1350 pennies.
Display the penny count as a dollar
std::cout << (pennyCount / 100) << "." << (pennyCount % 100) << std::endl;
I think this should look something like (untested)
13.50
Now your challenge is to make it look nice on the output.

Answer 3

The reason you only get 13 is because you are actually cutting off the least significant bits when you bit shift. Since you are cutting them off, there is no remainder to check. If you are interested in what your remainder is, you could do something like:
uint8_t x = 27;
Serial.println((x - (x >> 1) - (x >> 1));
(x - (x >> 1)) should give 14 here.
it would be pretty simple to add .5 to a number once you determine whether the remainder is 1.

Answer 4

The following should work and should be fast:
float y = (x >> 1) + (0.5 * (x & 0x01))
What it does
(x >> 1) Divide by 2 using the bit shift
(0.5 * (x & 0x01)) Add 0.5 if the last bit was 1 (odd number)