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I'm dealing with numbers, each one represented by two bytes (a short value, in the programming sense).

I want to compute the average of these two numbers, and get once again two bytes back.

Can I just do the following?

result_byte_A = first_num_byte_A + second_num_byte_A / 2
result_byte_B = first_num_byte_B + second_num_byte_B / 2

Will this yield a correct result? Or is this more complex than that?

Aviv Cohn
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1 Answers1

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Apart from the fact that you accidentally multiply instead of adding, there are overflow and rounding problems. Your method suffers from these in more cases than necessary. Try

short result = ((int) A + (int)B)/2;

instead. (This suffers only from rounding problems, e.g., the average of $4$ and $7$ is returned as $5$ instead of $5.5$, which cannot be expressed as short.

Your method (with + in place of *) goes awfully wrong, e.g. for $A=256$, $B=0$.

Hagen von Eitzen
  • 374,180