2020/2 winter

Question

Q:

Consider the following code snippet:

byte a, b;
...
byte average = (a + b) / 2;

The compiler complains “Required type: byte Provided: int” with respect to the second line. Why is that although only byte values are being involved?
Propose a solution solving the issue at the same preserving the byte result type. Is your solution susceptible to overflow errors?

Answer 1

A:

An expression “byte + byte” is of type int, see Binary operator type examples.
We require a cast:
```
...
byte average = (byte) ((a + b) / 2);
```
The expression a + b of type int may range from -2 * 128 to 2 * 127. Dividing by 2 leaves us with a byte's range. Thus arithmetic overflow errors cannot occur.

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