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riscv-gcc-1
Commit 6b148bd9 by Kazu Hirata Committed by Kazu Hirata
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crti.asm: Remove trailing spaces. 

	* config/h8300/crti.asm: Remove trailing spaces.
	* config/h8300/h8300.c: Likewise.
	* config/h8300/lib1funcs.asm: Likewise.

From-SVN: r53090
parent 609688f3
Showing with 57 additions and 51 deletions
gcc/ChangeLog
2002-05-03 Kazu Hirata <kazu@cs.umass.edu>
* config/h8300/crti.asm: Remove trailing spaces.
* config/h8300/h8300.c: Likewise.
* config/h8300/lib1funcs.asm: Likewise.
2002-05-02 Jason Merrill <jason@redhat.com>
* defaults.h (BOOL_TYPE_SIZE): Move default here from cp/decl.c.
......
gcc/config/h8300/crti.asm
/* Copyright (C) 2001 Free Software Foundation, Inc.
/* Copyright (C) 2001, 2002 Free Software Foundation, Inc.
This file is part of GNU CC.
......@@ -50,4 +50,4 @@ Boston, MA 02111-1307, USA. */
__init:
.section .fini
.global __fini
__fini:
__fini:
gcc/config/h8300/h8300.c
......@@ -2958,7 +2958,7 @@ compute_a_shift_length (insn, operands)
}
wlength += h8300_asm_insn_count (info.shift1) * n;
return 2 * wlength;
case SHIFT_ROT_AND:
......@@ -2977,7 +2977,7 @@ compute_a_shift_length (insn, operands)
}
wlength += h8300_asm_insn_count (info.shift1) * m;
/* Now mask off the high bits. */
switch (mode)
{
......
gcc/config/h8300/lib1funcs.asm
......@@ -2,7 +2,7 @@
;; Contributed by Steve Chamberlain <sac@cygnus.com>
;; Optimizations by Toshiyasu Morita <toshiyasu.morita@hsa.hitachi.com>
/* Copyright (C) 1994, 2000, 2001 Free Software Foundation, Inc.
/* Copyright (C) 1994, 2000, 2001, 2002 Free Software Foundation, Inc.
This file is free software; you can redistribute it and/or modify it
under the terms of the GNU General Public License as published by the
......@@ -162,7 +162,7 @@ ___ucmpsi2:
;; "supporting routines".
; general purpose normalize routine
;
;
; divisor in A0
; dividend in A1
; turns both into +ve numbers, and leaves what the answer sign
......@@ -174,13 +174,13 @@ ___ucmpsi2:
divnorm:
mov.b #0x0,A2L
or A0H,A0H ; is divisor > 0
bge _lab1
bge _lab1
not A0H ; no - then make it +ve
not A0L
adds #1,A0
adds #1,A0
xor #0x1,A2L ; and remember that in A2L
_lab1: or A1H,A1H ; look at dividend
bge _lab2
bge _lab2
not A1H ; it is -ve, make it positive
not A1L
adds #1,A1
......@@ -191,13 +191,13 @@ _lab2: rts
modnorm:
mov.b #0x0,A2L
or A0H,A0H ; is divisor > 0
bge _lab7
bge _lab7
not A0H ; no - then make it +ve
not A0L
adds #1,A0
adds #1,A0
xor #0x1,A2L ; and remember that in A2L
_lab7: or A1H,A1H ; look at dividend
bge _lab8
bge _lab8
not A1H ; it is -ve, make it positive
not A1L
adds #1,A1
......@@ -214,7 +214,7 @@ negans: or A2L,A2L ; should answer be negative ?
not A0H ; yes, so make it so
not A0L
adds #1,A0
_lab4: rts
_lab4: rts
; A0=A0%A1 signed
......@@ -251,13 +251,13 @@ ___umodhi3:
.global ___udivhi3
___udivhi3:
; A0 A1 A2 A3
; A0 A1 A2 A3
; Nn Dd P
sub.w A3,A3 ; Nn Dd xP 00
or A1H,A1H
sub.w A3,A3 ; Nn Dd xP 00
or A1H,A1H
bne divlongway
or A0H,A0H
beq _lab6
or A0H,A0H
beq _lab6
; we know that D == 0 and N is != 0
mov.b A0H,A3L ; Nn Dd xP 0N
......@@ -269,7 +269,7 @@ _lab6: mov.b A0L,A3L ; n
mov.b A3L,A0L ; Qq
mov.b A3H,A3L ; m
mov.b #0x0,A3H ; Qq 0m
rts
rts
; D != 0 - which means the denominator is
; loop around to get the result.
......@@ -280,19 +280,19 @@ divlongway:
mov.b #0x8,A2H ; 8
div8: add.b A0L,A0L ; n*=2
rotxl A3L ; Make remainder bigger
rotxl A3H
rotxl A3H
sub.w A1,A3 ; Q-=N
bhs setbit ; set a bit ?
add.w A1,A3 ; no : too far , Q+=N
dec A2H
bne div8 ; next bit
rts
dec A2H
bne div8 ; next bit
rts
setbit: inc A0L ; do insert bit
dec A2H
bne div8 ; next bit
rts
dec A2H
bne div8 ; next bit
rts
#endif /* __H8300__ */
#endif /* L_divhi3 */
......@@ -301,7 +301,7 @@ setbit: inc A0L ; do insert bit
;; 4 byte integer divides for the H8/300.
;;
;; We have one routine which does all the work and lots of
;; We have one routine which does all the work and lots of
;; little ones which prepare the args and massage the sign.
;; We bunch all of this into one object file since there are several
;; "supporting routines".
......@@ -334,11 +334,11 @@ divnorm:
postive:
mov.b A2H,A2H ; is the denominator -ve
bge postive2
not A2L
not A2L
not A2H
not A3L
not A3H
add.b #1,A3L
add.b #1,A3L
addx #0,A3H
addx #0,A2L
addx #0,A2H
......@@ -368,11 +368,11 @@ modnorm:
mpostive:
mov.b A2H,A2H ; is the denominator -ve
bge mpostive2
not A2L
not A2L
not A2H
not A3L
not A3H
add.b #1,A3L
add.b #1,A3L
addx #0,A3H
addx #0,A2L
addx #0,A2H
......@@ -424,7 +424,7 @@ mpostive2:
; denominator in A2/A3
.global ___modsi3
___modsi3:
PUSHP S2P
PUSHP S2P
PUSHP S0P
PUSHP S1P
......@@ -461,7 +461,7 @@ ___umodsi3:
mov.l S0P,A0P
#endif
bra exitdiv
.global ___divsi3
___divsi3:
PUSHP S2P
......@@ -477,7 +477,7 @@ exitdiv:
or S2L,S2L
beq reti
; should be -ve
#ifdef __H8300__
not A0H
......@@ -495,7 +495,7 @@ exitdiv:
reti:
POPP S2P
rts
rts
; takes A0/A1 numerator (A0P for 300H)
; A2/A3 denominator (A1P for 300H)
......@@ -538,7 +538,7 @@ NumByte3Zero:
mov.b S1H,S1L
mov.b #0x0,S1H
rts
rts
; have to do the divide by shift and test
DenHighZero:
......@@ -562,7 +562,7 @@ nextbit:
sub.w A3,S1 ; does it all fit
subx A2L,S0L
subx A2H,S0H
bhs setone
bhs setone
add.w A3,S1 ; no, restore mistake
addx A2L,S0L
......@@ -570,13 +570,13 @@ nextbit:
dec S2H
bne nextbit
rts
rts
setone:
inc A1L
dec S2H
bne nextbit
rts
rts
#else /* __H8300H__ */
......@@ -631,11 +631,11 @@ setone:
;; HImode multiply.
; The H8/300 only has an 8*8->16 multiply.
; The answer is the same as:
;
;
; product = (srca.l * srcb.l) + ((srca.h * srcb.l) + (srcb.h * srca.l)) * 256
; (we can ignore A1.h * A0.h cause that will all off the top)
; A0 in
; A1 in
; A1 in
; A0 answer
#ifdef __H8300__
......@@ -644,7 +644,7 @@ setone:
.global ___mulhi3
___mulhi3:
mov.b A1L,A2L ; A2l gets srcb.l
mulxu A0L,A2 ; A2 gets first sub product
mulxu A0L,A2 ; A2 gets first sub product
mov.b A0H,A3L ; prepare for
mulxu A1L,A3 ; second sub product
......@@ -652,7 +652,7 @@ ___mulhi3:
add.b A3L,A2H ; sum first two terms
mov.b A1H,A3L ; third sub product
mulxu A0L,A3
mulxu A0L,A3
add.b A3L,A2H ; almost there
mov.w A2,A0 ; that is
......@@ -664,7 +664,7 @@ ___mulhi3:
#ifdef L_mulsi3
;; SImode multiply.
;;
;;
;; I think that shift and add may be sufficient for this. Using the
;; supplied 8x8->16 would need 10 ops of 14 cycles each + overhead. This way
;; the inner loop uses maybe 20 cycles + overhead, but terminates
......@@ -673,7 +673,7 @@ ___mulhi3:
;; A0/A1 src_a
;; A2/A3 src_b
;;
;; while (a)
;; while (a)
;; {
;; if (a & 1)
;; r += b;
......@@ -691,10 +691,10 @@ ___mulsi3:
PUSHP S0P
PUSHP S1P
PUSHP S2P
sub.w S0,S0
sub.w S1,S1
; while (a)
_top: mov.w A0,A0
bne _more
......@@ -713,7 +713,7 @@ _nobit:
rotxr A0L
rotxr A1H
rotxr A1L
; b <<= 1
add.w A3,A3
addx A2L,A2L
......@@ -721,7 +721,7 @@ _nobit:
bra _top
_done:
mov.w S0,A0
mov.w S0,A0
mov.w S1,A1
POPP S2P
POPP S1P
......@@ -751,7 +751,7 @@ _done:
___mulsi3:
mov.w r1,r2 ; ( 2 states) b * d
mulxu r0,er2 ; (22 states)
mov.w e0,r3 ; ( 2 states) a * d
beq L_skip1 ; ( 4 states)
mulxu r1,er3 ; (22 states)
......
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