Loop Unrolling Fail in Simple Vectorized loop

Hi all,

Attached herewith is a simple vectorized function with loops performing a simple shuffle.

I want all loops (inner and outer) to be unrolled by 2 and as such used -unroll-count=2
The inner loops(with k as the induction variable and having constant trip counts) unroll fully, but the outer loop with (j) fails to unroll.

The llvm code is also attached with inner loops fully unrolled.

To inspect further, I added the following to the PassManagerBuilder.cpp to run some canonicalization routines and redo unrolling again. I have set partial unrolling on + have a huge threshold + allows expensive loop trip counts. Still it didn’t unroll by 2.




MPM.add(createLoopRotatePass(SizeLevel == 2 ? 0 : -1));


MPM.add(createIndVarSimplifyPass()); // Canonicalize indvars


Digging deeper I found, that it fails in UnrollRuntimeLoopRemainder function, where it is unable to calculate the BackEdge taken amount.

Can anybody explain what is need to get the outer loop unrolled by 2? It would be a great help.


test_unroll_sse2.c (1.4 KB)

out.ll (5.95 KB)

Well, I can at least explain what is happening… runtime unrolling needs to be able to symbolically compute the trip count to avoid inserting a branch after every iteration. SCEV isn’t able to prove that your loop isn’t an infinite loop (consider the case of vectorizable_elements==SIZE_MAX), therefore it can’t compute the trip count. Therefore, we don’t unroll. There’s a few different angles you could use to attack this: you could teach the unroller to unroll loops with an uncomputable trip count, or you can make the trip count of your loop computable somehow. Changing the unroller is probably straightforward (see the recently committed r284044). Making the trip count computable is more complicated… it’s probably possible to teach SCEV to reason about the overflow in the pointer computation, or maybe you could version the loop. -Eli

Thanks for the explanation. But I am a little confused with the following fact. Can’t LLVM keep vectorizable_elements as a symbolic value and convert the loop to say;

for(unsigned i = 0; i < vectorizable_elements ; i += 2){
//main loop

for(unsigned i=0 ; i < vectorizable_elements % 2; i++){
//fix up

Why does it have to reason about the range of vectorizable_elements? Even if vectorizable_elements == SIZE_MAX the above decomposition would work?

So, I tried unrolling the following simple loop.

int unroll(unsigned * a, unsigned * b, unsigned *c, unsigned count){

for(unsigned i=0; i<count; i++){

a[i] = b[i]*c[i-1];


return 0;


Then, the unroller is able to unroll it by 2 even though it doesn’t know about the range of count. SCEV of backedge taken count is (-1 + %count)

But, when I change the increment to 4, as in

int unroll(unsigned * a, unsigned * b, unsigned *c, unsigned count){

for(unsigned i=0; i<count; i+=4){

a[i] = b[i]*c[i-1];


return 0;


The unroller cannot compute the backedge taken count. Therefore, it seems like the problem is not with the range of “count”, can’t the unroller compute it as (- 1 + %count / 4)?

If count > MAX_UINT-4 your loop loops indefinitely with an increment of 4, I think.

Oh I see, the original loop may end normally, but by unrolling it may induce an infinite loop.

Oh I see, the original loop may end normally, but by unrolling it may
induce an infinite loop.

No. The problem is that step of the original loop is not 1.

  for(unsigned i=0; i<count; i+=4){
    a[i] = b[i]*c[i-1];

Let's assume unsigned is a 32 bit integer. Then maximum unsigned number
will be 2^32 - 1. Let count = 2^32 - 1. When the loop iterates at some
point we will have i = 2^32 - 4. What is the value of i in the next
iteration? 2^32 cannot be represented in a 32 bit integer, so what happens
is that you will wrap around and in the next iteration you will have i = 0.
So your original loop may be infinte. You can confirm this by compiling and
running the following program

#include <climits>
#include <iostream>

using namespace std;

int main() {

  for (int i = 0; i < UINT_MAX; i += 4) {
    if (i == 0)
      cout << "i == 0!" << endl;

  return 0;

In my example change the type of i (induction variable of the loop) to unsigned. It works, but when the type is “int” it can be controversial as int overflow is undefined behavior.