Question

在我介绍我正在使用的守则时,一个特别热点是:

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

阵列、a、x没有重叠。在我看来,像这种循环很容易成为病媒,然而,当我用“O3-ftree-vectorize-ftree-vectorizer-verbose=1”的“g++”汇编时,我没有发现这一特殊循环是病媒的。然而,在上述法典之前就发生了一起诉讼:

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

Answer 1

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.

Answer 2

一种可能性是,汇编者不能保证没有别的东西。换言之,汇编者如何确保<代码>y,a和x t不以某种方式重叠?

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