[GCC] PR target/86487: fix the way 'uses_hard_regs_p' handles paradoxical

subregs

gcc/ChangeLog:
2019-02-20 Andre Vieira  <andre.simoesdiasvieira@arm.com>

	PR target/86487
	* lra-constraints.c(uses_hard_regs_p): Fix handling of
	paradoxical SUBREGS.

gcc/testsuite/ChangeLog:
2019-02-20 Andre Vieira  <andre.simoesdiasvieira@arm.com>

	PR target/86487
	* gcc.target/arm/pr86487.c: New.

From-SVN: r269039

gcc/ChangeLog

+2019-02-20 Andre Vieira  <andre.simoesdiasvieira@arm.com>
+
+	PR target/86487
+	* lra-constraints.c(uses_hard_regs_p): Fix handling of
+	paradoxical SUBREGS.
+
 2019-02-20  Li Jia He  <helijia@linux.ibm.com>
 
 	PR target/88100

gcc/lra-constraints.c

@@ -1770,14 +1770,24 @@ uses_hard_regs_p (rtx x, HARD_REG_SET set)
     return false;
   code = GET_CODE (x);
   mode = GET_MODE (x);
+
   if (code == SUBREG)
     {
+      /* For all SUBREGs we want to check whether the full multi-register
+	 overlaps the set.  For normal SUBREGs this means 'get_hard_regno' of
+	 the inner register, for paradoxical SUBREGs this means the
+	 'get_hard_regno' of the full SUBREG and for complete SUBREGs either is
+	 fine.  Use the wider mode for all cases.  */
+      rtx subreg = SUBREG_REG (x);
       mode = wider_subreg_mode (x);
-      x = SUBREG_REG (x);
-      code = GET_CODE (x);
+      if (mode == GET_MODE (subreg))
+	{
+	  x = subreg;
+	  code = GET_CODE (x);
+	}
     }
 
-  if (REG_P (x))
+  if (REG_P (x) || SUBREG_P (x))
     {
       x_hard_regno = get_hard_regno (x, true);
       return (x_hard_regno >= 0

gcc/testsuite/ChangeLog

+2019-02-20 Andre Vieira  <andre.simoesdiasvieira@arm.com>
+
+	PR target/86487
+	* gcc.target/arm/pr86487.c: New.
+
 2019-02-20  Paolo Carlini  <paolo.carlini@oracle.com>
 
 	PR c++/84536

gcc/testsuite/gcc.target/arm/pr86487.c

+/* { dg-skip-if "" { *-*-* } { "-march=armv[0-6]*" "-mthumb" } { "" } } */
+/* { dg-require-effective-target arm_neon_hw } */
+/* { dg-options "-O1 -mbig-endian" } */
+/* { dg-add-options arm_neon } */
+int a, b, c, d;
+long long fn1(long long p2) { return p2 == 0 ? -1 : -1 % p2; }
+void fn2(long long p1, short p2, long p3) {
+  b = fn1((d || 6) & a);
+  c = b | p3;
+}