Re: RFA: Ada variable-sized objects, bit_size_type == TImode, and divti3

[kenner at vlsi1 dot ultra dot nyu dot edu (Richard Kenner)]

Sorry I forgot to get back to this when I got back last week.

    What appears to happen is that the Ada source in question defines a
    variable-sized object with non-standard alignment requirements, and
    Ada reflects this by building a DECL_SIZE_UNIT tree containing various
    mult_expr and div_expr nodes.  As alignment is counted in bits, those
    have type bit_size_type, which on 64-bit platforms corresponds to
    TImode. When expanding that tree, the backend emits calls to __divti3.

The point of having separate DECL_SIZE_UNIT and DECL_SIZE is precisely for
the former to only have computations in sizetype and the latter in bitsizetype.
There are very rare situations where a bitsizetype might be in the expression
for DECL_SIZE_UNIT, but other occurrences are a bug.  Indeed, there's
a trivial bug in layout_decl that would account for some of those usages.
Here's the diff; I'll do the testing and installation for it within the
next few days.

*************** layout_decl (decl, known_align)
*** 368,372 ****
        DECL_SIZE_UNIT (decl) = TYPE_SIZE_UNIT (type);
      }
!   else
      DECL_SIZE_UNIT (decl)
        = convert (sizetype, size_binop (CEIL_DIV_EXPR, DECL_SIZE (decl),
--- 365,369 ----
        DECL_SIZE_UNIT (decl) = TYPE_SIZE_UNIT (type);
      }
!   else if (DECL_SIZE_UNIT (decl) == 0)
      DECL_SIZE_UNIT (decl)
        = convert (sizetype, size_binop (CEIL_DIV_EXPR, DECL_SIZE (decl),


    - does bit_size_type really need to be TImode

Perhaps not, but it should for consistency.  In practice, not that much needs
to be computed dynamically for bitsizes.

    - if so, shouldn't these special cases be implemented more efficiently
      somewhere (we divide by an integer constant 8 here)

No.

    - in general, are we really supposed to need TImode division

Yes.

    How does this work on other 64-bit platforms?

By having __divti3.