Loop unrolling in clang

Question

I am trying to selectively unroll the second loop in the following program:

#include <stdio.h>

int main()
{
    int in[1000], out[1000]; 
    int i,j;

    #pragma nounroll
    for (i = 100; i < 1000; i++)
    {
       in[i]+= 10;
    }

    #pragma unroll 2
    for (j = 100; j < 1000; j++)
    {
       out[j]+= 10;
    }

    return 1;
}

When I run clang (3.5) with the following options, it unrolls both the loops 4 times.

clang -std=c++11 -O3 -fno-slp-vectorize -fno-vectorize -mllvm -unroll-count=4 -mllvm -debug-pass=Arguments -emit-llvm -c *.cpp 

What am I doing wrong? Also, if I add -fno-unroll-loops, or skip the -unroll-count=4 flag, it does not unroll any loop.

Also, any hints on how to debug pragma errors?

Answer 1

I think there is no support for such pragmas in clang 3.5.

However starting from 3.6, you can use #pragma clang loop unroll(enable | disable) to enable or disable the automatic diagnostics-based unroll feature. If you want to fully unroll a loop then #pragma clang loop unroll(full) is a shorthand for that. You can also use #pragma clang loop unroll_count(N) - where N is a compile-time constant - to explicitly specify the unroll count.

More info here.

Your code rewritten in terms of the above stuff:

#include <stdio.h>

int main()
{

  int in[1000], out[1000]; 
  int i,j;

  #pragma clang loop unroll(disable)
  for (i = 100; i < 1000; i++)
  {
     in[i]+= 10;
  }

  #pragma clang loop unroll_count(2)
  for (j = 100; j < 1000; j++)
  {
     out[j]+= 10;
  }


  return 1;
}

Answer 2

-unroll-count=4 has a higher priority than #pragma clang loop unroll_count(2). That's why it ends up unroll it by 4. Meaning the compiler is following the unroll-count command line option NOT the pragma. Also as plasmacel mentioned, #pragma clang loop unroll is not supported before clang 3.6.