Correct ix86_rtx_cost for multi-word multiplication.

This is the i386 backend specific piece of my revised patch for
PR middle-end/98865, where Richard Biener has suggested that I perform
the desired transformation during RTL expansion where the backend can
control whether it is profitable to convert a multiplication into a
bit-wise AND and a negation.  This works well for x86_64, but alas
exposes a latent bug with -m32, where a DImode multiplication incorrectly
appears to be cheaper than negdi2+anddi3(!?).  The fix to ix86_rtx_costs
is to report that a DImode (multi-word) multiplication actually requires
three SImode multiplications and two SImode additions.  This also corrects
the cost of TImode multiplication on TARGET_64BIT.

2022-05-18  Roger Sayle  <roger@nextmovesoftware.com>

gcc/ChangeLog
	* config/i386/i386.cc (ix86_rtx_costs) [MULT]: When mode size
	is wider than word_mode, a multiplication costs three word_mode
	multiplications and two word_mode additions.

diff --git a/gcc/config/i386/i386.cc b/gcc/config/i386/i386.cc
index 86752a6..30a9cd0 100644 (file)
--- a/gcc/config/i386/i386.cc
+++ b/gcc/config/i386/i386.cc
@@ -20634,7 +20634,17 @@ ix86_rtx_costs (rtx x, machine_mode mode, int outer_code_i, int opno,
                op0 = XEXP (op0, 0), mode = GET_MODE (op0);
            }
 
-         *total = (cost->mult_init[MODE_INDEX (mode)]
+         int mult_init;
+         // Double word multiplication requires 3 mults and 2 adds.
+         if (GET_MODE_SIZE (mode) > UNITS_PER_WORD)
+           {
+             mult_init = 3 * cost->mult_init[MODE_INDEX (word_mode)]
+                         + 2 * cost->add;
+             nbits *= 3;
+           }
+         else mult_init = cost->mult_init[MODE_INDEX (mode)];
+
+         *total = (mult_init
                    + nbits * cost->mult_bit
                    + rtx_cost (op0, mode, outer_code, opno, speed)
                    + rtx_cost (op1, mode, outer_code, opno, speed));