Question

Let’s say I am given integers x and y (satisfying x <= y with ones digit of 0 so they are, in particular, divisible by two). Then I know that their average avg = ((x+y) / 2) is an integer as well. I would like to find this midpoint rounded up to a resolution of 100. In other words if my two inputs are 75200 and 75300 then the avg is 75250 and rounded up to the nearest 100 (but without exceeding or equaling the bigger number) forces the answer to be 75200.

How can I implement this logic without first dividing everything by 100 and using the following floating point arithmetic:

x + std::floor((y - x) * .5 * 100 + .5)*0.01

In other words, how can I do the above without floating point values but obtain the same behavior at the resolution of 100 instead of 0.01?

Answer 1

To compute the average you can do

avg = (x + y) / 2

(BTW, integer addition and division by 2 are very cheap operations even on small microcontrollers.)

To round this to the nearest multiple of 100 (corresponding to your floating-point example) you can do

result = ((avg + 50) / 100) * 100

as integer division rounds down to the nearest integer. By changing the 50 to 0 you can always round down, while changing it to 99 always rounds up.

Edit: Note that this method for rounding doesn’t work for negative numbers. Since integer division rounds towards zero, in that case you’ll need to subtract the 50, subtract 99 to always round down and subtract 0 to always round up.