return 0;
}
+static int
+walk_mems_2 (rtx *x, rtx mem)
+{
+ if (MEM_P (*x))
+ {
+ if (alias_sets_conflict_p (MEM_ALIAS_SET(*x), MEM_ALIAS_SET(mem)))
+ return 1;
+
+ return -1;
+ }
+ return 0;
+}
+
+static int
+walk_mems_1 (rtx *x, rtx *pat)
+{
+ if (MEM_P (*x))
+ {
+ /* Visit all MEMs in *PAT and check indepedence. */
+ if (for_each_rtx (pat, (rtx_function) walk_mems_2, *x))
+ /* Indicate that dependence was determined and stop traversal. */
+ return 1;
+
+ return -1;
+ }
+ return 0;
+}
+
+/* Return 1 if two specified instructions have mem expr with conflict alias sets*/
+bool
+insn_alias_sets_conflict_p (rtx insn1, rtx insn2)
+{
+ /* For each pair of MEMs in INSN1 and INSN2 check their independence. */
+ return for_each_rtx (&PATTERN (insn1), (rtx_function) walk_mems_1,
+ &PATTERN (insn2));
+}
+
/* Return 1 if the two specified alias sets will always conflict. */
int
}
/* Variant qualifiers don't affect the alias set, so get the main
- variant. If this is a type with a known alias set, return it. */
+ variant. Always use the canonical type as well.
+ If this is a type with a known alias set, return it. */
t = TYPE_MAIN_VARIANT (t);
+ if (TYPE_CANONICAL (t))
+ t = TYPE_CANONICAL (t);
if (TYPE_ALIAS_SET_KNOWN_P (t))
return TYPE_ALIAS_SET (t);
{
unsigned int regno;
+#if defined (FIND_BASE_TERM)
+ /* Try machine-dependent ways to find the base term. */
+ src = FIND_BASE_TERM (src);
+#endif
+
switch (GET_CODE (src))
{
case SYMBOL_REF:
return 0;
/* Fall through. */
case LO_SUM:
+ /* The standard form is (lo_sum reg sym) so look only at the
+ second operand. */
+ return find_base_term (XEXP (x, 1));
case PLUS:
case MINUS:
{
if (rtx_equal_p (x_base, y_base))
return 1;
- /* The base addresses of the read and write are different expressions.
- If they are both symbols and they are not accessed via AND, there is
- no conflict. We can bring knowledge of object alignment into play
- here. For example, on alpha, "char a, b;" can alias one another,
- though "char a; long b;" cannot. */
+ /* The base addresses are different expressions. If they are not accessed
+ via AND, there is no conflict. We can bring knowledge of object
+ alignment into play here. For example, on alpha, "char a, b;" can
+ alias one another, though "char a; long b;" cannot. AND addesses may
+ implicitly alias surrounding objects; i.e. unaligned access in DImode
+ via AND address can alias all surrounding object types except those
+ with aligment 8 or higher. */
+ if (GET_CODE (x) == AND && GET_CODE (y) == AND)
+ return 1;
+ if (GET_CODE (x) == AND
+ && (GET_CODE (XEXP (x, 1)) != CONST_INT
+ || (int) GET_MODE_UNIT_SIZE (y_mode) < -INTVAL (XEXP (x, 1))))
+ return 1;
+ if (GET_CODE (y) == AND
+ && (GET_CODE (XEXP (y, 1)) != CONST_INT
+ || (int) GET_MODE_UNIT_SIZE (x_mode) < -INTVAL (XEXP (y, 1))))
+ return 1;
+
+ /* Differing symbols not accessed via AND never alias. */
if (GET_CODE (x_base) != ADDRESS && GET_CODE (y_base) != ADDRESS)
- {
- if (GET_CODE (x) == AND && GET_CODE (y) == AND)
- return 1;
- if (GET_CODE (x) == AND
- && (GET_CODE (XEXP (x, 1)) != CONST_INT
- || (int) GET_MODE_UNIT_SIZE (y_mode) < -INTVAL (XEXP (x, 1))))
- return 1;
- if (GET_CODE (y) == AND
- && (GET_CODE (XEXP (y, 1)) != CONST_INT
- || (int) GET_MODE_UNIT_SIZE (x_mode) < -INTVAL (XEXP (y, 1))))
- return 1;
- /* Differing symbols never alias. */
- return 0;
- }
+ return 0;
/* If one address is a stack reference there can be no alias:
stack references using different base registers do not alias,