I think that you’re interested in a more general tutorial on how to vectorize code. I’m sorry I’m going to plug in my own articles  , but I’ve just tried to write a short version here and I basically almost rewrote them so there was
no point. For your case, I’d read at least the first half of “Residual Iterations” from Part 1 and most of part 2. The concept of residual iterations and reductions is something you’ll see over and over and it happens
in your example too. Let me explain the use of shufflevector pretty quick and then I’ll explain some specifics in your example. Also, before we move any further, I recommend that we turn off loop unrolling in your
example because it doesn’t add educational value and just clutters the code, so here’s your updated version .
— Shufflevector —
If you are not sure whether you understand what the instruction alone does, let me know. You can also play with this LLVM IR . I recommend that you run this code and change the mask in line 24.
The reason it’s used in your example (see ) is because the compiler wants to create a vector which has
arr (referring to the C++ entity) in all its lanes. In line 4 (see ) it’s loaded into %0, in line 17 it inserts
it only to the 0th (first) lane of a vector (and for the time being, we don’t care about the other lanes which is why the first arg to
poison) and in line 18, it shuffles the value. The shuffle mask is the vector [0, 0, 0, 0]
(in short, zeroinitializer) which means in the 0th lane of the result vector (%minmax.ident.splat) insert the value of the 0th lane of the input vector (%minmax.ident.splatinsert). In the 1st lane of the result
vector, insert the value of the 0th lane of the input vector etc. So, all lanes of the result vector get the 0th lane of the input vector.
— Residual Iterations —
Assuming that you have understood the use of residual iterations, let’s see where it happens in your example. First of all, notice that in
entry, there is a check whether
n is bigger than 1, because if it’s not, there are no
iterations to do at all and in this case, the result, i.e.,
arr, which is why it’s loaded so early. Check the phi at line 45. Then, if we make it to
for.body.preheader, there is a check whether we have at least
4 iterations (the fact that this loop starts from one makes the computations somewhat more complicated but not a lot). Because if we don’t, then there’s no point going into the vectorized version. Then, skipping what the vectorized
code does exactly for now (but check how the iteration count is rounded down to the nearest multiple of 4 in line 15), check that after the vectorized code, the flow moves to
which essentially checks whether there are any more iterations to do and if so, jumps to the serial version of the loop (this is where the residual iterations happen).
— Reduction —
In the article, I explain reductions using accumulation as an example, but in your example there’s another reduction, namely, finding the maximum element. Notice that this is also a reduction: new_value = max_of(old_value, data).
The reduction variable (acting as both old and new value) is
max in your example. The operation is max_of and the continuously new data is arr[i].
Finding the maximum element is a commutative operation. For instance, to find the maximum element among 4, you can split them in two pairs any way you want, find the maximum of each pair independently
of the other pair (those two maximums are partial results and since their computation is independent of each other, it can be done in parallel), and finally find the maximum among these two maximums.
What happens in the vectorized code is that each lane finds a partial maximum. In line 27 it loads 4 elements and in the next two lines, it “compares” them with the previous 4 elements (%vec.phi, which keeps the 4 partial maximums which are updated
in every iteration). It computes a vector which has 4 maximums, one for each pair of lanes (between the new 4 elements in %wide.load, in %wide.load, and %vec.phi). You might want to run an example in paper to see the effect. Consider that you want to compute
the maximum of 9 numbers, in this order: 2, 10, 7, 3, 1, 8, 9, 11, 4
First, you get 2 (arr) and broadcast it in a vector: <2, 2, 2, 2>. This is the starting value of %vec.phi. Then, you load 4 values: %wide.load = <10, 7, 3, 1>
Then, you do an element-wise comparison of the two and for each pair of lanes, you keep the maximum. In this case, 10 > 2, 7 > 2, 3 > 2 but 1 < 2. So, %5 = <10, 7, 3, 2>, which becomes the new %vec.phi.
Next iteration, you load another 4: %wide.load = <8, 9, 11, 4> and you compare with (the new) %vec.phi: 8 < 10, 9 > 7, 11 > 3, 4 > 2. So, %5 = <10, 9, 11, 4>. Now, you have computed 4 partial maximums
and you only need to find the maximum among those 4 to find the total maximum. That happens in line 35 and you’re done.
Hope this helps! Let me know if there are any questions.
Στις Δευ, 7 Ιουν 2021 στις 8:49 π.μ., ο/η Craig Topper <firstname.lastname@example.org> έγραψε: