DCPU-16 DIV instruction

Question

I'm looking at the specification for the DCPU-16 and I'm having trouble understanding the purpose of the overflow value with the DIV instruction: DIV a, b - sets a to a/b, sets O to ((a<<16)/b)&0xffff. Can anybody explain the semantic meaning of O here, what it would be useful for?

Accepted Answer

It looks like O gives the fractional part of the result (as a fixed-point number). For example, consider 5 / 2:

```
a = 5 / 2 = 2 (integer part)
O = ((5 << 16) / 2) & 0xffff = (327680 / 2) & 0xffff = 32768
```

If you consider O as the 16 binary fractional digits, then this represents the 0.5 fractional part of the result.

Another way of looking at it is the binary result in bits is:

```
aaaaaaaaaaaaaaaa.OOOOOOOOOOOOOOOO
```

5 / 2 is

```
0000000000000010.1000000000000000
```

As you can tell by inspection, that result is 5 (101 binary) shifted right by one bit into the fractional bits.

DCPU-16 DIV instruction

Solution

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