How does Duff's device work?

Question

I've read the article on Wikipedia on the Duff's device, and I don't get it. I am really interested, but I've read the explanation there a couple of times and I still don't get it how the Duff's device works.

What would a more detailed explanation be?

Answer 1

Just experimenting, found another variant getting along without interleaving switch and loop:

int n = (count + 1) / 8;
switch (count % 8)
{
    LOOP:
case 0:
    if(n-- == 0)
        break;
    putchar('.');
case 7:
    putchar('.');
case 6:
    putchar('.');
case 5:
    putchar('.');
case 4:
    putchar('.');
case 3:
    putchar('.');
case 2:
    putchar('.');
case 1:
    putchar('.');
default:
    goto LOOP;
}

Answer 2

There are two key things to Duff's device. First, which I suspect is the easier part to understand, the loop is unrolled. This trades larger code size for more speed by avoiding some of the overhead involved in checking whether the loop is finished and jumping back to the top of the loop. The CPU can run faster when it's executing straight-line code instead of jumping.

The second aspect is the switch statement. It allows the code to jump into the middle of the loop the first time through. The surprising part to most people is that such a thing is allowed. Well, it's allowed. Execution starts at the calculated case label, and then it falls through to each successive assignment statement, just like any other switch statement. After the last case label, execution reaches the bottom of the loop, at which point it jumps back to the top. The top of the loop is inside the switch statement, so the switch is not re-evaluated anymore.

The original loop is unwound eight times, so the number of iterations is divided by eight. If the number of bytes to be copied isn't a multiple of eight, then there are some bytes left over. Most algorithms that copy blocks of bytes at a time will handle the remainder bytes at the end, but Duff's device handles them at the beginning. The function calculates count % 8 for the switch statement to figure what the remainder will be, jumps to the case label for that many bytes, and copies them. Then the loop continues to copy groups of eight bytes.

Answer 3

Though I'm not 100% sure what you're asking for, here goes...

The issue that Duff's device addresses is one of loop unwinding (as you'll no doubt have seen on the Wiki link you posted). What this basically equates to is an optimisation of run-time efficiency, over memory footprint. Duff's device deals with serial copying, rather than just any old problem, but is a classic example of how optimisations can be made by reducing the number of times that a comparison needs to be done in a loop.

As an alternative example, which may make it easier to understand, imagine you have an array of items you wish to loop over, and add 1 to them each time... ordinarily, you might use a for loop, and loop around 100 times. This seems fairly logical and, it is... however, an optimisation can be made by unwinding the loop (obviously not too far... or you may as well just not use the loop).

So a regular for loop:

for(int i = 0; i < 100; i++)
{
    myArray[i] += 1;
}

becomes

for(int i = 0; i < 100; i+10)
{
    myArray[i] += 1;
    myArray[i+1] += 1;
    myArray[i+2] += 1;
    myArray[i+3] += 1;
    myArray[i+4] += 1;
    myArray[i+5] += 1;
    myArray[i+6] += 1;
    myArray[i+7] += 1;
    myArray[i+8] += 1;
    myArray[i+9] += 1;
}

What Duff's device does is implement this idea, in C, but (as you saw on the Wiki) with serial copies. What you're seeing above, with the unwound example, is 10 comparisons compared to 100 in the original - this amounts to a minor, but possibly significant, optimisation.