Why isn't this loop vectorized?

Question

One particular hot spot when I profile a code I am working on, is the following loop:

for(int loc = start; loc<end; ++loc)
    y[loc]+=a[offset+loc]*x[loc+d];

where the arrays y, a, and x have no overlap. It seems to me that a loop like this should be easily vectorized, however when I compile using g++ with the options "-O3 -ftree-vectorize -ftree-vectorizer-verbose=1", I get no indication that this particular loop was vectorized. However, a loop occurring just before the code above:

for(int i=0; i<m; ++i)
    y[i]=0;

does get vectorized according to the output. Any thoughts on why the first loop is not vectorized, or how I might be able to fix this? (I am not all that educated on the concept of vectorization, so I am likely missing something quite obvious)

As per Oli's suggestion, turning up the verbosity yields the following notes (while I am usually good at reading compiler warnings/errors/output, I have no idea what this means):

./include/mv_ops.h:89: note: dependence distance  = 0.
./include/mv_ops.h:89: note: accesses have the same alignment.
./include/mv_ops.h:89: note: dependence distance modulo vf == 0 between *D.50620_89 and *D.50620_89
./include/mv_ops.h:89: note: not vectorized: can't determine dependence between *D.50623_98 and *D.50620_89

Why don't you work with pointers to a[offset] and x[d] instead? — K-ballo
Thanks for the suggestion K-ballo. Just did what you suggested, which makes things a bit more readable (though it has no effect on my vectorizing). — MarkD
Are those x, y, and a declared in that function as arrays or pointers? — MSN
@Oli Charlesworth: Its just a matter of personal preference. That's why it was a comment and not an answer. — K-ballo

MSN MSN · Accepted Answer · 2011-09-22T01:20:23

You need to tell the compiler that x, y, and a do not overlap. In C/C++ terms that means telling the compiler that those pointers do not alias by declaring them with restrict (or __restrict). gcc is very aggressive about optimizations when it assumes no aliasing, so be careful.

Why isn't this loop vectorized?

2 Answers