diff options
Diffstat (limited to 'gcc/cse.c')
| -rw-r--r-- | gcc/cse.c | 18 |
1 files changed, 5 insertions, 13 deletions
diff --git a/gcc/cse.c b/gcc/cse.c index 6880787f4c5..15e2dc8e60b 100644 --- a/gcc/cse.c +++ b/gcc/cse.c @@ -1461,13 +1461,7 @@ insert (rtx x, struct table_elt *classp, unsigned int hash, enum machine_mode mo elt->related_value = 0; elt->in_memory = 0; elt->mode = mode; - elt->is_const = (CONSTANT_P (x) - /* GNU C++ takes advantage of this for `this' - (and other const values). */ - || (REG_P (x) - && RTX_UNCHANGING_P (x) - && REGNO (x) >= FIRST_PSEUDO_REGISTER) - || fixed_base_plus_p (x)); + elt->is_const = (CONSTANT_P (x) || fixed_base_plus_p (x)); if (table[hash]) table[hash]->prev_same_hash = elt; @@ -2094,7 +2088,7 @@ canon_hash_string (const char *ps) Store 1 in do_not_record if any subexpression is volatile. Store 1 in hash_arg_in_memory if X contains a MEM rtx - which does not have the RTX_UNCHANGING_P bit set. + which does not have the MEM_READONLY_P bit set. Note that cse_insn knows that the hash code of a MEM expression is just (int) MEM plus the hash code of the address. */ @@ -2229,7 +2223,7 @@ canon_hash (rtx x, enum machine_mode mode) do_not_record = 1; return 0; } - if (! RTX_UNCHANGING_P (x) || fixed_base_plus_p (XEXP (x, 0))) + if (!MEM_READONLY_P (x)) hash_arg_in_memory = 1; /* Now that we have already found this special case, @@ -2249,7 +2243,7 @@ canon_hash (rtx x, enum machine_mode mode) hash += (unsigned) USE; x = XEXP (x, 0); - if (! RTX_UNCHANGING_P (x) || fixed_base_plus_p (XEXP (x, 0))) + if (!MEM_READONLY_P (x)) hash_arg_in_memory = 1; /* Now that we have already found this special case, @@ -5974,9 +5968,7 @@ cse_insn (rtx insn, rtx libcall_insn) sets[i].dest_hash, GET_MODE (dest)); elt->in_memory = (MEM_P (sets[i].inner_dest) - && (! RTX_UNCHANGING_P (sets[i].inner_dest) - || fixed_base_plus_p (XEXP (sets[i].inner_dest, - 0)))); + && !MEM_READONLY_P (sets[i].inner_dest)); /* If we have (set (subreg:m1 (reg:m2 foo) 0) (bar:m1)), M1 is no narrower than M2, and both M1 and M2 are the same number of words, |