On 21 October 2011 16:25, Jakub Jelinek <ja...@redhat.com> wrote: > On Fri, Oct 21, 2011 at 03:44:11PM +0200, Ira Rosen wrote: >> But it's OK to allow modifier != NONE if it's not SLP, so we need &&, no? > > Well, in my patch that check was guarded by the if (slp_node ...), > so presumably it would allow modifier == NARROW vectorization in the loops > (otherwise some testcases would fail I'd hope).
The problem with that is that slp_node can be NULL but it can still be an SLP stmt (as you probably have guessed judging by the following questions ;)) > > Is gcc_assert ((slp_node != NULL) == (STMT_SLP_TYPE (stmt_info) != 0)); > always true? No. > If not, when it is not? STMT_SLP_TYPE (stmt_info) != 0 may mean HYBRID_SLP_STMT, meaning that we are vectorizing the stmt both as SLP and as regular loop vectorization. So in the regular loop transformation of a hybrid stmt (STMT_SLP_TYPE (stmt_info) != 0) doesn't (entail slp_node != NULL). The other direction is always true. > When would be slp_node == NULL > and PURE_SLP_STMT true? In the analysis of loop SLP. In loop SLP we analyze all the stmts of the loop in their original order (and not as in BB SLP where we just analyze SLP nodes). A stmt can belong to more than one SLP node, and we may also need to vectorize it in a regular loop-vectorization way at the same time. So, during the analysis we don't have stmt's SLP node. (Note that during the analysis we need to know ncopies only to verify that the operation is supported and for cost estimation). And this is another case when 'if (STMT_SLP_TYPE (stmt_info) != 0) then (slp_node != NULL)' is false. I hope this makes sense. Ira > > Jakub >
