[Bug tree-optimization/101925] [10/11/12 Regression] reversed storage order when compiling with -O3 only since r10-4742-g9b75f56d4b7951c6
rguenth at gcc dot gnu.org
gcc-bugzilla@gcc.gnu.org
Mon Aug 16 11:20:41 GMT 2021
https://gcc.gnu.org/bugzilla/show_bug.cgi?id=101925
Richard Biener <rguenth at gcc dot gnu.org> changed:
What |Removed |Added
----------------------------------------------------------------------------
Assignee|rguenth at gcc dot gnu.org |unassigned at gcc dot gnu.org
CC| |jamborm at gcc dot gnu.org
Status|ASSIGNED |NEW
--- Comment #6 from Richard Biener <rguenth at gcc dot gnu.org> ---
Interestingly when I use
struct _be_ip6_addr {
union {
short addr8[16];
} BIG_ENDIAN u;
} BIG_ENDIAN;
(note 'short' instead of 'char') I _do_ get
rc.u.addr8[0] {rev}= _2;
...
rc.u.addr8[1] {rev}= _4;
...
so either the testcase is bogus or I fail to capture the desired semantics
here. We do end up with aggregate copies here after inlining:
rc.u.addr6 = D.2017;
(note also no reverse storage marker). But then SRA comes along and
scalarizes that aggregate copy:
<bb 3> :
_3 = ip.u.addr;
- rc.u.addr {rev}= _3;
+ rc$u$addr_152 = _3;
goto <bb 5>; [INV]
...
+ rc.u.addr6.u.addr8[0] = SR.48_91;
+ rc.u.addr6.u.addr8[1] = SR.49_92;
+ rc.u.addr6.u.addr8[2] = SR.50_93;
+ rc.u.addr6.u.addr8[3] = SR.51_94;
+ rc.u.addr6.u.addr8[4] = SR.52_95;
+ rc.u.addr6.u.addr8[5] = SR.53_96;
+ rc.u.addr6.u.addr8[6] = SR.54_97;
+ rc.u.addr6.u.addr8[7] = SR.55_98;
+ rc.u.addr6.u.addr8[8] = SR.56_99;
+ rc.u.addr6.u.addr8[9] = SR.57_100;
+ rc.u.addr6.u.addr8[10] = SR.58_101;
+ rc.u.addr6.u.addr8[11] = SR.59_102;
+ rc.u.addr6.u.addr8[12] = SR.60_103;
+ rc.u.addr6.u.addr8[13] = SR.61_104;
+ rc.u.addr6.u.addr8[14] = SR.62_105;
+ rc.u.addr6.u.addr8[15] = SR.63_106;
+ rc$u$addr_123 ={rev} MEM <int32_t> [(union *)&rc + 4B];
(??)
...
+ rc.is_v4 {rev}= rc$is_v4_74;
+ rc.u.addr {rev}= rc$u$addr_8;
+ MEM <char> [(struct _be_net_addr *)&rc + 8B] = rc$u$addr6$u$addr8$4_47;
+ MEM <char> [(struct _be_net_addr *)&rc + 9B] = rc$u$addr6$u$addr8$5_48;
+ MEM <char> [(struct _be_net_addr *)&rc + 10B] = rc$u$addr6$u$addr8$6_49;
+ MEM <char> [(struct _be_net_addr *)&rc + 11B] = rc$u$addr6$u$addr8$7_50;
+ MEM <char> [(struct _be_net_addr *)&rc + 12B] = rc$u$addr6$u$addr8$8_51;
+ MEM <char> [(struct _be_net_addr *)&rc + 13B] = rc$u$addr6$u$addr8$9_52;
+ MEM <char> [(struct _be_net_addr *)&rc + 14B] = rc$u$addr6$u$addr8$10_53;
+ MEM <char> [(struct _be_net_addr *)&rc + 15B] = rc$u$addr6$u$addr8$11_54;
+ MEM <char> [(struct _be_net_addr *)&rc + 16B] = rc$u$addr6$u$addr8$12_55;
+ MEM <char> [(struct _be_net_addr *)&rc + 17B] = rc$u$addr6$u$addr8$13_56;
+ MEM <char> [(struct _be_net_addr *)&rc + 18B] = rc$u$addr6$u$addr8$14_57;
+ MEM <char> [(struct _be_net_addr *)&rc + 19B] = rc$u$addr6$u$addr8$15_58;
<retval> = rc;
I'm somewhat lost as to how this works.
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