My brain hurts. I'm supposed to know this stuff.

[Bunk]

Here's another question - why are you doing this?

It beats murdering HR people.

Or does it?

[zeitghost]

What Doodab said. y is accumulating the unused lower order bits so that every now and again it'll add one to the higher order bits that are used. Without it, a stream of 0x0080s would only ever send zero to the hardware and the signal is discarded; with it you get alternating 0x00 and 0x01s.

A sort of dithery thing then.

Interesting.

[zeitghost]

Here's another question - why are you doing this?

It marginally relieves the tedium.

[Platypus]

Here's another question - why are you doing this?

[zeitghost]

One assumes that the caller is putting whatever data needs to be written in the high order byte.

I still can't get my head around this.

Here's another question:

Problem 3—What is the difference between an
interpreter and a compiler?

Slight difference in level, what?

Similarly:

Problem 1—Why is it a generally a bad idea to
write lengthy ISRs?

[VectraMan]

What Doodab said. y is accumulating the unused lower order bits so that every now and again it'll add one to the higher order bits that are used. Without it, a stream of 0x0080s would only ever send zero to the hardware and the signal is discarded; with it you get alternating 0x00 and 0x01s.

[zeitghost]

Yes, but why?

If the DAC is connected to bits 8 to 15, wtf does y have to do with owt?

Some of those EQ questions are mind bending.

But only if you don't know the answer, like.

Plus there's no way to find the answer now since the question was posed about 13 years ago.

[Platypus]

Ah got it. Because it's an 8 bit DAC, you maintain the lower at bits at 0x8000 and only change the upper bits.

Badly explained but I know what I mean

[stek]

Why don't you just tippex it out?

Sometime you have to think around a problem...

[doodab]

y appears to be keeping track of the low order part of the result of cumulative addition of x to itself. Presumably something to do with keeping the result accurate over many iterations because the value of x is truncated when it's written to the DAC?

[Platypus]

Weird, y remembers the last value of x therefore you can switch bits on but not switch them off again, since y is added to current x.

[zeitghost]

Problem 3—An embedded 16-bit microcontroller is connected to an 8-bit DAC, where the DAC is connected to
the high-order half of the data bus.

What is the purpose of y in the following driver?

#define DAC (*(unsigned *) 0x8000)
void DAC_out (unsigned x)
{
static unsigned y;
x += y;
y = x & 0x00FF;
DAC = x;
}

WTF?

I haven't a clue.

It's part of an EQ test out of circuit cellar issue 119.