Hi,
when two arrays of scalars have a different storage order in Ada, the
front-end makes sure that the conversion is performed component-wise
so that each component can be reversed. So it's a little bit counter
productive that the ldist pass performs the opposite transformation
and synthesizes a memcpy/memmove in this case.
Tested on x86-64/Linux, OK for the mainline?
2022-11-22 Eric Botcazou
* tree-loop-distribution.cc (loop_distribution::classify_builtin_ldst):
Bail out if source and destination do not have the same storage order.
2022-11-22 Eric Botcazou
* gnat.dg/sso18.adb: New test.
--
Eric Botcazoudiff --git a/gcc/tree-loop-distribution.cc b/gcc/tree-loop-distribution.cc
index ed3dd73e1a9..15ae2410861 100644
--- a/gcc/tree-loop-distribution.cc
+++ b/gcc/tree-loop-distribution.cc
@@ -1790,10 +1790,15 @@ loop_distribution::classify_builtin_ldst (loop_p loop, struct graph *rdg,
if (res != 2)
return;
- /* They much have the same access size. */
+ /* They must have the same access size. */
if (!operand_equal_p (size, src_size, 0))
return;
+ /* They must have the same storage order. */
+ if (reverse_storage_order_for_component_p (DR_REF (dst_dr))
+ != reverse_storage_order_for_component_p (DR_REF (src_dr)))
+return;
+
/* Load and store in loop nest must access memory in the same way, i.e,
their must have the same steps in each loop of the nest. */
if (dst_steps.length () != src_steps.length ())
-- { dg-do run }
-- { dg-options "-O2" }
with System;
procedure SSO18 is
type Arr is array (1..32) of Short_Integer;
type Rev_Arr is array (1..32) of Short_Integer
with Scalar_Storage_Order => System.High_Order_First;
C : constant Arr := (others => 16);
RA : Rev_Arr;
A : Arr;
begin
RA := Rev_Arr(C);
A := Arr (RA);
if A /= C or else RA(1) /= 16 then
raise Program_Error;
end if;
end;
