On Thu, Jun 13, 2019 at 09:50:16AM -0600, Martin Sebor wrote:
> On 6/13/19 9:34 AM, Jakub Jelinek wrote:
> > On Thu, Jun 13, 2019 at 09:30:37AM -0600, Martin Sebor wrote:
> > > The size of the access above doesn't look right. The test is:
> >
> > It is correct. You have MEM <int[5]> [(int *)&i], which is
> > the same thing as i itself, and on this you apply an ARRAY_REF,
> > which is printed after it, with index j_1(D). ARRAY_REF is applied on
> > arrays and the result type is the array element type, so int in this case.
>
> Aah, it's two REFs in one. I misread the array index as being
> a part of the MEM_REF operand, like this:
>
> MEM <int[5]> [((int *)&i)[j_1(D)]] = 1;
>
> I guess I've never noticed this before. Why is the whole thing
> not simplified to an ARRAY_REF?
>
> i[j_2(D)] = 1;
No idea in this case, though of course there can be other cases e.g.
where the MEM_REF has different number of elements, different element type
etc. from the underlying variable or where the MEM_REF first operand is not
address, but pointer.
Jakub
