Hello, > > >>> with base being either some memory object or an INDIRECT_REF of a > > >>> pointer and be done with that tree code. > > >> So if you have MEM_REF(INDIRECT_REF(a),i,0), you really haven't done > > >> any better in removing recursion :) > > > > > > What type the first operand would be could be a one-bit flag in the > > > MEM_REF itself. I.e. if there's an implicit INDIRECT_REF around the > > > first operand, or not. The important part is only that the target > > > address of that memory reference is computable completely trivially, > > > namely by A + i, where A is either a or &a depending on that flag. > > > And the more important thing anyway is that alias information of this > > > specific mem reference encoded therein. > > > > Not for data dependence it's not. :) > > You mean tests based on the actual index vector, instead of the final > offset vector? I still sometimes think that one should be able to > recover the index vector reasonably well, when given only offsets (and > perhaps adjusted bases). It's a tradeoff of having only one way to > express memory accesses (which I find quite sexy) and having to do a > bit more work when trying to get at the offsets.
by this kind of lowering, you lose quite a lot of information. For example, a[i][j] lowers to something like offset = step1 * i + step2 * j; MEM(a, offset). 1) Recovering the index vector is not easy, as the optimizations may change this code a lot. Even if you find an expression that looks like MEM(a, step1 * i + step2 * j), you have no good way to know that this was not originally a[i + 1][j - step1/step2]. 2) If "a" has constant bounds, before the lowering you could use these bounds for example to estimate # of iterations of a loop. From similar reasons as in the last statement, this is impossible after lowering. 3) There may be some uses of the fact that we know that the arithmetics in step1 * i + step2 * j cannot overflow (which is at the moment impossible to remember after lowering), although I cannot think about a good example just now. Zdenek
