【发布时间】:2016-08-28 21:10:36
【问题描述】:
我有一个看起来像这样的热点。在这里收集某种矢量会很好......关于如何让编译器喜欢这个的任何建议?
do ii = 1, N
if (diff(ii) .le. M ) then
i = i0 + ii - 1
rbuf( irb ) = i
irb = irb + 1
end if
end do
使用 ifort 16.0.2 我的 opt 报告看起来像
LOOP BEGIN at code.f(197,13)
remark #25084: Preprocess Loopnests: Moving Out Store [ code.f(203,13) ]
remark #15344: loop was not vectorized: vector dependence prevents vectorization
remark #15346: vector dependence: assumed FLOW dependence between irb line 201 and irb line 200
remark #15346: vector dependence: assumed ANTI dependence between irb line 200 and irb line 201
remark #15346: vector dependence: assumed ANTI dependence between irb line 200 and irb line 201
remark #15346: vector dependence: assumed FLOW dependence between irb line 201 and irb line 200
remark #25439: unrolled with remainder by 2
remark #25015: Estimate of max trip count of loop=1600
LOOP END
这里是小测试程序
program vect
integer :: ii, i0, irb
integer, parameter :: N=32
integer, parameter :: M=8
integer, dimension(N) :: diff
integer, dimension(2*N) :: rbuf
rbuf = 0
!only some values of diff will meet condition
!could be random
do ii=1, N
diff(ii) = ii
end do
!from an outer loop
i0=1003
!this is code for filling up a buffer for an expensive vectorized
!subroutine with full vectors, irb < 2*N
irb=3
do ii = 1, N
if (diff(ii) .le. M ) then
i = i0 + ii - 1
rbuf( irb ) = i
irb = irb + 1
end if
end do
!check
do ii = 1, 2*N
write(*,*) ii, rbuf(ii)
end do
end
【问题讨论】:
-
diff是函数还是数组?你能创建一个最小的可编译示例来玩吗? -
@AlexanderVogt 现在举个例子
-
您正在复制一个数组,根据条件过滤掉一些元素?请参阅this question,以及 SSE 和 AVX2 答案(以及 AVX512 答案)。向量化是可能的,但 C 编译器不会为你做这件事。可能也不是 Fortran 编译器。您将需要与 C 内在函数等效的东西,或者只调用 C 或 asm 函数。
标签: fortran vectorization intel simd