;; Copyright (C) 2001-2021 Free Software Foundation, Inc. ;; ;; This file is part of GCC. ;; ;; GCC is free software; you can redistribute it and/or modify it under ;; the terms of the GNU General Public License as published by the Free ;; Software Foundation; either version 3, or (at your option) any later ;; version. ;; ;; GCC is distributed in the hope that it will be useful, but WITHOUT ANY ;; WARRANTY; without even the implied warranty of MERCHANTABILITY or ;; FITNESS FOR A PARTICULAR PURPOSE. See the GNU General Public License ;; for more details. ;; ;; Under Section 7 of GPL version 3, you are granted additional ;; permissions described in the GCC Runtime Library Exception, version ;; 3.1, as published by the Free Software Foundation. ;; ;; You should have received a copy of the GNU General Public License and ;; a copy of the GCC Runtime Library Exception along with this program; ;; see the files COPYING3 and COPYING.RUNTIME respectively. If not, see ;; . ;; ;; This code is derived from mulsi3.S, observing that the mstep*16-based ;; multiplications there, from which it is formed, are actually ;; zero-extending; in gcc-speak "umulhisi3". The difference to *this* ;; function is just a missing top mstep*16 sequence and shifts and 64-bit ;; additions for the high part. Compared to an implementation based on ;; calling __Mul four times (see default implementation of umul_ppmm in ;; longlong.h), this will complete in a time between a fourth and a third ;; of that, assuming the value-based optimizations don't strike. If they ;; all strike there (very often) but none here, we still win, though by a ;; lesser margin, due to lesser total overhead. #define L(x) .x #define CONCAT1(a, b) CONCAT2(a, b) #define CONCAT2(a, b) a ## b #ifdef __USER_LABEL_PREFIX__ # define SYM(x) CONCAT1 (__USER_LABEL_PREFIX__, x) #else # define SYM(x) x #endif .global SYM(__umulsidi3) .type SYM(__umulsidi3),@function SYM(__umulsidi3): #if defined (__CRIS_arch_version) && __CRIS_arch_version >= 10 ;; Can't have the mulu.d last on a cache-line, due to a hardware bug. See ;; the documentation for -mmul-bug-workaround. ;; Not worthwhile to conditionalize here. .p2alignw 2,0x050f mulu.d $r11,$r10 ret move $mof,$r11 #else move.d $r11,$r9 bound.d $r10,$r9 cmpu.w 65535,$r9 bls L(L3) move.d $r10,$r12 move.d $r10,$r13 movu.w $r11,$r9 ; ab*cd = (a*c)<<32 (a*d + b*c)<<16 + b*d ;; We're called for floating point numbers very often with the "low" 16 ;; bits zero, so it's worthwhile to optimize for that. beq L(L6) ; d == 0? lslq 16,$r13 beq L(L7) ; b == 0? clear.w $r10 mstep $r9,$r13 ; d*b mstep $r9,$r13 mstep $r9,$r13 mstep $r9,$r13 mstep $r9,$r13 mstep $r9,$r13 mstep $r9,$r13 mstep $r9,$r13 mstep $r9,$r13 mstep $r9,$r13 mstep $r9,$r13 mstep $r9,$r13 mstep $r9,$r13 mstep $r9,$r13 mstep $r9,$r13 mstep $r9,$r13 L(L7): test.d $r10 mstep $r9,$r10 ; d*a mstep $r9,$r10 mstep $r9,$r10 mstep $r9,$r10 mstep $r9,$r10 mstep $r9,$r10 mstep $r9,$r10 mstep $r9,$r10 mstep $r9,$r10 mstep $r9,$r10 mstep $r9,$r10 mstep $r9,$r10 mstep $r9,$r10 mstep $r9,$r10 mstep $r9,$r10 mstep $r9,$r10 ;; d*a in $r10, d*b in $r13, ab in $r12 and cd in $r11 ;; $r9 = d, need to do b*c and a*c; we can drop d. ;; so $r9 is up for use and we can shift down $r11 as the mstep ;; source for the next mstep-part. L(L8): lsrq 16,$r11 move.d $r12,$r9 lslq 16,$r9 beq L(L9) ; b == 0? mstep $r11,$r9 mstep $r11,$r9 ; b*c mstep $r11,$r9 mstep $r11,$r9 mstep $r11,$r9 mstep $r11,$r9 mstep $r11,$r9 mstep $r11,$r9 mstep $r11,$r9 mstep $r11,$r9 mstep $r11,$r9 mstep $r11,$r9 mstep $r11,$r9 mstep $r11,$r9 mstep $r11,$r9 mstep $r11,$r9 L(L9): ;; d*a in $r10, d*b in $r13, c*b in $r9, ab in $r12 and c in $r11, ;; need to do a*c. We want that to end up in $r11, so we shift up $r11 to ;; now use as the destination operand. We'd need a test insn to update N ;; to do it the other way round. lsrq 16,$r12 lslq 16,$r11 mstep $r12,$r11 mstep $r12,$r11 mstep $r12,$r11 mstep $r12,$r11 mstep $r12,$r11 mstep $r12,$r11 mstep $r12,$r11 mstep $r12,$r11 mstep $r12,$r11 mstep $r12,$r11 mstep $r12,$r11 mstep $r12,$r11 mstep $r12,$r11 mstep $r12,$r11 mstep $r12,$r11 mstep $r12,$r11 ;; d*a in $r10, d*b in $r13, c*b in $r9, a*c in $r11 ($r12 free). ;; Need (a*d + b*c)<<16 + b*d into $r10 and ;; a*c + (a*d + b*c)>>16 plus carry from the additions into $r11. add.d $r9,$r10 ; (a*d + b*c) - may produce a carry. scs $r12 ; The carry corresponds to bit 16 of $r11. lslq 16,$r12 add.d $r12,$r11 ; $r11 = a*c + carry from (a*d + b*c). #if defined (__CRIS_arch_version) && __CRIS_arch_version >= 8 swapw $r10 addu.w $r10,$r11 ; $r11 = a*c + (a*d + b*c) >> 16 including carry. clear.w $r10 ; $r10 = (a*d + b*c) << 16 #else move.d $r10,$r9 lsrq 16,$r9 add.d $r9,$r11 ; $r11 = a*c + (a*d + b*c) >> 16 including carry. lslq 16,$r10 ; $r10 = (a*d + b*c) << 16 #endif add.d $r13,$r10 ; $r10 = (a*d + b*c) << 16 + b*d - may produce a carry. scs $r9 ret add.d $r9,$r11 ; Last carry added to the high-order 32 bits. L(L6): clear.d $r13 ba L(L8) clear.d $r10 L(L11): clear.d $r10 ret clear.d $r11 L(L3): ;; Form the maximum in $r10, by knowing the minimum, $r9. ;; (We don't know which one of $r10 or $r11 it is.) ;; Check if the largest operand is still just 16 bits. xor $r9,$r10 xor $r11,$r10 cmpu.w 65535,$r10 bls L(L5) movu.w $r9,$r13 ;; We have ab*cd = (a*c)<<32 + (a*d + b*c)<<16 + b*d, but c==0 ;; so we only need (a*d)<<16 + b*d with d = $r13, ab = $r10. ;; Remember that the upper part of (a*d)<<16 goes into the lower part ;; of $r11 and there may be a carry from adding the low 32 parts. beq L(L11) ; d == 0? move.d $r10,$r9 lslq 16,$r9 beq L(L10) ; b == 0? clear.w $r10 mstep $r13,$r9 ; b*d mstep $r13,$r9 mstep $r13,$r9 mstep $r13,$r9 mstep $r13,$r9 mstep $r13,$r9 mstep $r13,$r9 mstep $r13,$r9 mstep $r13,$r9 mstep $r13,$r9 mstep $r13,$r9 mstep $r13,$r9 mstep $r13,$r9 mstep $r13,$r9 mstep $r13,$r9 mstep $r13,$r9 L(L10): test.d $r10 mstep $r13,$r10 ; a*d mstep $r13,$r10 mstep $r13,$r10 mstep $r13,$r10 mstep $r13,$r10 mstep $r13,$r10 mstep $r13,$r10 mstep $r13,$r10 mstep $r13,$r10 mstep $r13,$r10 mstep $r13,$r10 mstep $r13,$r10 mstep $r13,$r10 mstep $r13,$r10 mstep $r13,$r10 mstep $r13,$r10 move.d $r10,$r11 lsrq 16,$r11 lslq 16,$r10 add.d $r9,$r10 scs $r12 ret add.d $r12,$r11 L(L5): ;; We have ab*cd = (a*c)<<32 + (a*d + b*c)<<16 + b*d, but a and c==0 ;; so b*d (with min=b=$r13, max=d=$r10) it is. As it won't overflow the ;; 32-bit part, just set $r11 to 0. lslq 16,$r10 clear.d $r11 mstep $r13,$r10 mstep $r13,$r10 mstep $r13,$r10 mstep $r13,$r10 mstep $r13,$r10 mstep $r13,$r10 mstep $r13,$r10 mstep $r13,$r10 mstep $r13,$r10 mstep $r13,$r10 mstep $r13,$r10 mstep $r13,$r10 mstep $r13,$r10 mstep $r13,$r10 mstep $r13,$r10 ret mstep $r13,$r10 #endif L(Lfe1): .size SYM(__umulsidi3),L(Lfe1)-SYM(__umulsidi3)