问题
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?
回答1:
There are some good explanations elsewhere, but let me give it a try. (This is a lot easier on a whiteboard!) Here's the Wikipedia example with some notations.
Let's say you're copying 20 bytes. The flow control of the program for the first pass is:
int count; // Set to 20
{
int n = (count + 7) / 8; // n is now 3. (The "while" is going
// to be run three times.)
switch (count % 8) { // The remainder is 4 (20 modulo 8) so
// jump to the case 4
case 0: // [skipped]
do { // [skipped]
*to = *from++; // [skipped]
case 7: *to = *from++; // [skipped]
case 6: *to = *from++; // [skipped]
case 5: *to = *from++; // [skipped]
case 4: *to = *from++; // Start here. Copy 1 byte (total 1)
case 3: *to = *from++; // Copy 1 byte (total 2)
case 2: *to = *from++; // Copy 1 byte (total 3)
case 1: *to = *from++; // Copy 1 byte (total 4)
} while (--n > 0); // N = 3 Reduce N by 1, then jump up
// to the "do" if it's still
} // greater than 0 (and it is)
}
Now, start the second pass, we run just the indicated code:
int count; //
{
int n = (count + 7) / 8; //
//
switch (count % 8) { //
//
case 0: //
do { // The while jumps to here.
*to = *from++; // Copy 1 byte (total 5)
case 7: *to = *from++; // Copy 1 byte (total 6)
case 6: *to = *from++; // Copy 1 byte (total 7)
case 5: *to = *from++; // Copy 1 byte (total 8)
case 4: *to = *from++; // Copy 1 byte (total 9)
case 3: *to = *from++; // Copy 1 byte (total 10)
case 2: *to = *from++; // Copy 1 byte (total 11)
case 1: *to = *from++; // Copy 1 byte (total 12)
} while (--n > 0); // N = 2 Reduce N by 1, then jump up
// to the "do" if it's still
} // greater than 0 (and it is)
}
Now, start the third pass:
int count; //
{
int n = (count + 7) / 8; //
//
switch (count % 8) { //
//
case 0: //
do { // The while jumps to here.
*to = *from++; // Copy 1 byte (total 13)
case 7: *to = *from++; // Copy 1 byte (total 14)
case 6: *to = *from++; // Copy 1 byte (total 15)
case 5: *to = *from++; // Copy 1 byte (total 16)
case 4: *to = *from++; // Copy 1 byte (total 17)
case 3: *to = *from++; // Copy 1 byte (total 18)
case 2: *to = *from++; // Copy 1 byte (total 19)
case 1: *to = *from++; // Copy 1 byte (total 20)
} while (--n > 0); // N = 1 Reduce N by 1, then jump up
// to the "do" if it's still
} // greater than 0 (and it's not, so bail)
} // continue here...
20 bytes are now copied.
Note: The original Duff's Device (shown above) copied to an I/O device at the to address. Thus, it wasn't necessary to increment the pointer *to. When copying between two memory buffers you'd need to use *to++.
回答2:
The explanation in Dr. Dobb's Journal is the best that I found on the topic.
This being my AHA moment:
for (i = 0; i < len; ++i) {
HAL_IO_PORT = *pSource++;
}
becomes:
int n = len / 8;
for (i = 0; i < n; ++i) {
HAL_IO_PORT = *pSource++;
HAL_IO_PORT = *pSource++;
HAL_IO_PORT = *pSource++;
HAL_IO_PORT = *pSource++;
HAL_IO_PORT = *pSource++;
HAL_IO_PORT = *pSource++;
HAL_IO_PORT = *pSource++;
HAL_IO_PORT = *pSource++;
}
n = len % 8;
for (i = 0; i < n; ++i) {
HAL_IO_PORT = *pSource++;
}
becomes:
int n = (len + 8 - 1) / 8;
switch (len % 8) {
case 0: do { HAL_IO_PORT = *pSource++;
case 7: HAL_IO_PORT = *pSource++;
case 6: HAL_IO_PORT = *pSource++;
case 5: HAL_IO_PORT = *pSource++;
case 4: HAL_IO_PORT = *pSource++;
case 3: HAL_IO_PORT = *pSource++;
case 2: HAL_IO_PORT = *pSource++;
case 1: HAL_IO_PORT = *pSource++;
} while (--n > 0);
}
回答3:
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.
回答4:
The point of duffs device is to reduce the number of comparisons done in a tight memcpy implementation.
Suppose you want to copy 'count' bytes from a to b, the straight forward approach is to do the following:
do {
*a = *b++;
} while (--count > 0);
How many times do you need to compare count to see if it's a above 0? 'count' times.
Now, the duff device uses a nasty unintentional side effect of a switch case which allows you to reduce the number of comparisons needed to count / 8.
Now suppose you want to copy 20 bytes using duffs device, how many comparisons would you need? Only 3, since you copy eight bytes at a time except the last first one where you copy just 4.
UPDATED: You don't have to do 8 comparisons/case-in-switch statements, but it's reasonable a trade-off between function size and speed.
回答5:
When I read it for the first time, I autoformatted it to this
void dsend(char* to, char* from, count) {
int n = (count + 7) / 8;
switch (count % 8) {
case 0: do {
*to = *from++;
case 7: *to = *from++;
case 6: *to = *from++;
case 5: *to = *from++;
case 4: *to = *from++;
case 3: *to = *from++;
case 2: *to = *from++;
case 1: *to = *from++;
} while (--n > 0);
}
}
and I had no idea what was happening.
Maybe not when this question was asked, but now Wikipedia has a very good explanation
The device is valid, legal C by virtue of two attributes in C:
- Relaxed specification of the switch statement in the language's definition. At the time of the device's invention this was the first edition of The C Programming Language which requires only that the controlled statement of the switch be a syntactically valid (compound) statement within which case labels can appear prefixing any sub-statement. In conjunction with the fact that, in the absence of a break statement, the flow of control will fall-through from a statement controlled by one case label to that controlled by the next, this means that the code specifies a succession of count copies from sequential source addresses to the memory-mapped output port.
- The ability to legally jump into the middle of a loop in C.
回答6:
1: Duffs device is a particular implimentation of loop unrolling. What is loop unrolling?
If you have an operation to perform N times in a loop you can trade program size for speed by executing the loop N/n times and then in the loop inlining (unrolling) the loop code n times e.g. replacing:
for (int i=0; i<N; i++) {
// [The loop code...]
}
with
for (int i=0; i<N/n; i++) {
// [The loop code...]
// [The loop code...]
// [The loop code...]
...
// [The loop code...] // n times!
}
Which works great if N % n == 0 - no need for Duff! If that is not true then you have to handle the remainder - which is a pain.
2: How does Duffs device differ from this standard loop unrolling?
Duffs device is just a clever way of dealing with the remainder loop cycles when N % n != 0. The whole do / while executes N / n number of times as per standard loop unrolling (because the case 0 applies). On the last run through the loop (the 'N/n+1'th time) the case kicks in and we jump to the N % n case and run the loop code the 'remainder' number of times.
回答7:
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.
回答8:
Here's a non-detailed explanation which is what I feel to be the crux of Duff's device:
The thing is, C is basically a nice facade for assembly language (PDP-7 assembly to be specific; if you studied that you would see how striking the similarities are). And, in assembly language, you don't really have loops - you have labels and conditional-branch instructions. So the loop is just a part of the overall sequence of instructions with a label and a branch somewhere:
instruction
label1: instruction
instruction
instruction
instruction
jump to label1 some condition
and a switch instruction is branching/jumping ahead somewhat:
evaluate expression into register r
compare r with first case value
branch to first case label if equal
compare r with second case value
branch to second case label if equal
etc....
first_case_label:
instruction
instruction
second_case_label:
instruction
instruction
etc...
In assembly it's easily conceivable how to combine these two control structures, and when you think of it that way, their combination in C doesn't seem so weird anymore.
回答9:
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;
}
来源：https://stackoverflow.com/questions/514118/how-does-duffs-device-work