loop unrolling

Accelerated Computing CUDA CUDA Programming and Performance
1

I have following code using loop unrolling:

#pragma unroll

for (int i=0;i<n;i++)

{

    ....

}

here if n is a defined constant, everything works fine. However, if n is a variable, performance dramatically reduced. I noticed roughly 3 times the instructions are issued and executed. Can anyone justify this? Alternative solution is welcomed as well.

2

With n constant, is it possible that the compiler can pre-compute the result of some or all of the result of the loop? That sort of optimization would explain the performance difference.

3

what is occupancy?

could you list register usage by “nvcc -Xptxas -v -arch=sm_20 [source file]”?

Usually, compiler would use more registers to do loop unrolling.
By the way, if n is a constant, then compiler would unroll the loop automatically.

4

My intention is to use loop unrolling to improve performance. I guess if it’s a compiler level technique, then I won’t be able to change the iteration cycle at run time.

5

Occupancy is fine. I checked with visual profiler. The number of instructions issued increases 3 times, comparing to unrolled case. I am just looking for ways to do loop unrolling at run time. May be that’s just not feasible.

6

Hi There, Small Potato!

The compiler un-rolls even if “n” is not a compile time constant…
BUt you need to give the parameter for “unrolling”…, I think…
Try “#pragma unroll 5”… The compiler will generate code that will divide “n” by 5, and then un-roll appropriately.
You can check the PTX… That should tell u clearly what the compiler is doing…

7

In my finite difference kernels for wave equations (very similar to FDTD3d from SDK) unrolling increases the performance.
However, manual loop unrolling produces even faster code: ptxas reports smaller register usage if I unroll manually compared to #pragma unroll

8

Loop unrolling happens at compile time.