Shift Double Precision

Were you asking me I had no idea why Intel has support for shift double precision in the 80×86. Probably their answer would be “because it used to be a CISC processor”. The shift double precision is pretty easy to implement algorithm. But maybe it was popular back then and they decided to support it hardware-ly. Like now that they add very important instructions to the SSE sets. Even so, everyone (includes me) seems to implement the algorithm like this:

(a << c) | (b >> (32-c))

Where a and b are the 32 bits input variables(/registers) and c is the count. The code shows a shift left double precision. Shifting right will require to change the shifts direction for each one of the shifts. However, if a and b were 16 bits, the equation of the second shift amount changes to (16-c). And now there is a problem, why? Because we might enter into the magical world of undefined behavior. And why is that? Because the first thing that describes the shift/rotate instructions is that the count operand is masked to preserve only the 5 least significant bits. This is because the largest shift amount for a 32 bits input is 32 shifts (and then you get a 0, ignore SAR for now). And if the input is 16 bits, the count is still masked with 31. That means that you can shift a 16 bits register more than its size. Which doesn’t make much sense, but possible for other shift instructions. But when you use a shift double preicision, not that it doesn’t makes sense, it is also undefined. That is the result is undefined, because then you try to move bits from b into a. But the count becomes negative. For example: shld ax, bx, 17. And internally the second shift amount is calculated as (16-c) which becomes (16-17). And that’s bad, right?

In reality everything is defined when it comes to digital logic. Even the undefined stuff. There must be a reason to the result I get from executing such an instruction like in the example above, even though it’s correctly and officially undefined. And I know that there is a rational behind it, because the result is consistent (at least to my Intel Core2Duo processor). So being the stubborn I am, I decided I want to know how that calculation is really being done in the hardware level.

I forgot to mention that the reason I care of how to implement this instruction is because I have to simulate it for the Vial project. I guess eventually it’s a waste of time, but I really wanted to know what’s going on anyway. Therefore I decided to research the matter and get with the algorithm my processor uses. Examining the results of officially undefined results, I quickly managed to see how to calculate the shift like the processor does, and it goes like this for 16 bits input (I guess, it will work the same for 8 bits input as well, and note that 32 bits input can’t have an undefined range, because you can’t get a negative shift amount):

def shld(a, b, c):
 c &= 31
 if c <= 15:
  return ((a << c) | (b >> (16-c))) & 0xffff

  # Undefined behavior:
  c &= 15
  return ((b << c) | (a >> (16-c))) & 0xffff

Yes, the code is in Python. But you can see that if the the count is bigger than 15, then we are replacing the input order. And then comes the part where you say “NOW WTF?!”. Even though I got this algorithm to return the same results as the processor does for defined and undefined input, I could wager the processor won’t do this kind of stuff internally. So I sat down some (long) more, and stared at the code, doing a few experiments here and there. Eventually it occurred to me:

def shld(a, b, c):
 c &= 31
 x = a | (b << 16)
 return ((x << c) | (x >> (32-c))) & 0xffff

Now you can see that the input for the original equation is the same bits-buffer input, which contains both inputs together as one. Taking a count of 17, won’t yield a negative register, but something else. Anyway, I have no idea why they implemented this instruction like they did (and it applies to SHRD as well), but I believe it has something to do with the way their processor so-called ‘engine’ works and hardware stuff.

After I learned how it works I was so eager to see how it works on AMD. And guess what? They don’t work the same, where it comes to the undefined behavior, of course. And since I don’t have an AMD anymore I didn’t see how they really implemented their shift double precision instructions.

In the Vial project, where I simulate these instructions, I added a special check for the count, to see that it’s not bigger than the input size, and if it is, I mark the destination register and some of the flags as Undefined. This way I will know when I do code-analysis that something is really wrong/buggy with the way the application works. Now what if the application is purposely uses the undefined behavior? Screw us both then. Now why would a sane application do that? ohh and that’s another story…

By the way, other shift/rotate instructions don’t have any problem with the shift amount since they can’t yield negative shift amount internally in any way, therefore the results are always defined for every input.

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