Off and on, I've been working on a user-only target of hppa to qemu. It's now
about 95% working. If anyone would like to try it out, it's available at
git://github.com/rth7680/qemu.git tgt-hppa
While implementing the unit-type instructions, I wondered why no one (outside
hp?) had written a version of the string routines utilizing the UXOR insn, with
the SomeByteZero and NoByteZero conditions.
Attached are versions of strlen, strchr and strrchr. They pass simple tests
within my emulator; I'd be interested to know if they pass full glibc testing
on real hardware.
Thanks,
r~
;! HP-PA strlen
;! Copyright (C) 2016 Free Software Foundation, Inc.
.text
.export strlen
.balign 16
strlen:
.proc
.callinfo frame=0,no_calls
.entry
;! Compute the number of bytes required to align the pointer.
;! Shifting by 1 gets us 4 insns to play with per entry.
ldo -1(%r26), %r20
depw,z %r20, 30, 2, %r20
blr %r20, %r0
copy %r26, %r28
;! ptr % 4 == 1
ldb 0(%r26), %r20
cmpiclr,= 0, %r20, %r0
b,n 9f
ldo 1(%r26), %r26
;! ptr % 4 == 2
ldb 0(%r26), %r20
cmpiclr,= 0, %r20, %r0
b,n 9f
ldo 1(%r26), %r26
;! ptr % 4 == 3
ldb 0(%r26), %r20
cmpiclr,= 0, %r20, %r0
b,n 9f
ldo 1(%r26), %r26
;! ptr % 4 == 0
;! Main loop. Use the Some Byte Zero unit condition to find
;; a word containing the string terminator.
0: ldw,ma 4(%r26), %r20
uxor,sbz %r0, %r20, %r0
b,n 0b
;! Found, somewhere in with word in %r20. Test each byte in
;! sequence, computing the appopriate offset from %r26 into %r21.
ldo -1(%r26), %r21
extrw,u,<> %r20, 23, 8, %r0
ldo -2(%r26), %r21
extrw,u,<> %r20, 15, 8, %r0
ldo -3(%r26), %r21
extrw,u,<> %r20, 7, 8, %r0
ldo -4(%r26), %r21
bv 0(%r2)
sub %r21, %r28, %r28
;! Found, with no displacement off %r26.
9: bv 0(%r2)
sub %r26, %r28, %r28
.exit
.procend
;! HP-PA strchr
;! Copyright (C) 2016 Free Software Foundation, Inc.
.text
.export strchr
.balign 16
strchr:
.proc
.callinfo frame=0,no_calls
.entry
;! Compute the number of bytes required to align the pointer.
;! Multiply by 3, giving us 6 insns per entry to work with.
ldo -1(%r26), %r20
extrw,u %r25, 31, 8, %r25
extrw,u %r20, 31, 2, %r20
shladd,l %r20, 1, %r20, %r20
blr %r20, %r0
copy %r26, %r28
;! ptr % 4 == 1
ldb 0(%r28), %r20
cmpclr,<> %r25, %r20, %r0
bv,n 0(%r2)
cmpclr,<> %r0, %r20, %r0
b,n 9f
ldo 1(%r28), %r28
;! ptr % 4 == 2
ldb 0(%r28), %r20
cmpclr,<> %r25, %r20, %r0
bv,n 0(%r2)
cmpclr,<> %r0, %r20, %r0
b,n 9f
ldo 1(%r28), %r28
;! ptr % 4 == 3
ldb 0(%r28), %r20
cmpclr,<> %r25, %r20, %r0
bv,n 0(%r2)
cmpclr,<> %r0, %r20, %r0
b,n 9f
ldo 1(%r28), %r28
;! ptr % 4 == 0
ldw,ma 4(%r28), %r20
depw %r25, 23, 8, %r25
depw %r25, 15, 16, %r25
;! Main loop. Use the No Byte Zero unit condition to find
;; a word containing C or 0.
0: uxor,nbz %r25, %r20, %r0
b,n 1f
uxor,nbz %r0, %r20, %r0
b,n 1f
b 0b
ldw,ma 4(%r28), %r20
;! Found, somewhere in with word in %r20.
;! Test each byte in sequence.
1: extrw,u %r25, 31, 8, %r25
extrw,u %r20, 7, 8, %r21
ldo -4(%r28), %r28
cmpclr,<> %r25, %r21, %r0
bv,n 0(%r2)
cmpclr,<> %r0, %r21, %r0
b,n 9f
extrw,u %r20, 15, 8, %r21
ldo 1(%r28), %r28
cmpclr,<> %r25, %r21, %r0
bv,n 0(%r2)
cmpclr,<> %r0, %r21, %r0
b,n 9f
extrw,u %r20, 23, 8, %r21
ldo 1(%r28), %r28
cmpclr,<> %r25, %r21, %r0
bv,n 0(%r2)
cmpclr,<> %r0, %r21, %r0
b,n 9f
extrw,u %r20, 31, 8, %r21
ldo 1(%r28), %r28
cmpclr,<> %r25, %r21, %r0
bv,n 0(%r2)
;! String terminator found before the search character.
9: bv 0(%r2)
ldi 0, %r28
.exit
.procend
;! HP-PA strrchr
;! Copyright (C) 2016 Free Software Foundation, Inc.
.text
.export strrchr
.balign 16
strrchr:
.proc
.callinfo frame=0,no_calls
.entry
;! Compute the number of bytes required to align the pointer.
;! Multiply by 3, giving us 6 insns per entry to work with.
ldo -1(%r26), %r20
extrw,u %r25, 31, 8, %r25
extrw,u %r20, 31, 2, %r20
shladd,l %r20, 1, %r20, %r20
blr %r20, %r0
;! Begin by assuming that C is not present.
ldi 0, %r28
;! ptr % 4 == 1
ldb 0(%r26), %r20
cmpclr,<> %r25, %r20, %r0
copy %r26, %r28
cmpclr,<> %r0, %r20, %r0
bv,n 0(%r2)
ldo 1(%r26), %r26
;! ptr % 4 == 2
ldb 0(%r26), %r20
cmpclr,<> %r25, %r20, %r0
copy %r26, %r28
cmpclr,<> %r0, %r20, %r0
bv,n 0(%r2)
ldo 1(%r26), %r26
;! ptr % 4 == 3
ldb 0(%r26), %r20
cmpclr,<> %r25, %r20, %r0
copy %r26, %r28
cmpclr,<> %r0, %r20, %r0
bv,n 0(%r2)
ldo 1(%r26), %r26
;! ptr % 4 == 0
ldw,ma 4(%r26), %r20
copy %r25, %r24
depw %r24, 23, 8, %r24
depw %r24, 15, 16, %r24
;! Main loop.
0: ;; Test for a NUL terminator within the word and exit if found.
uxor,nbz %r0, %r20, %r0
b,n 1f
;; Test for C within the word. If not found, loop and load the
;; next word in the delay slot. If found, load the next word
;; now anyway, since we know that we havn't seen end-of-string.
copy %r20, %r21
uxor,sbz %r24, %r20, %r0
b 0b
ldw,ma 4(%r26), %r20
;; Found C within the "current" word. Note that it is now in %r21,
;; and the address for the beginning of that word is now -8(%r26),
;; since we have incremented the pointer twice since the load.
extrw,u %r21, 7, 8, %r22
cmpclr,<> %r25, %r22, %r0
ldo -8(%r26), %r28
extrw,u %r21, 15, 8, %r22
cmpclr,<> %r25, %r22, %r0
ldo -7(%r26), %r28
extrw,u %r21, 23, 8, %r22
cmpclr,<> %r25, %r22, %r0
ldo -6(%r26), %r28
extrw,u %r21, 31, 8, %r22
cmpclr,<> %r25, %r22, %r0
ldo -5(%r26), %r28
b,n 0b
;! Found NUL somewhere in with word in %r20, loaded from -4(%r26).
;! Test each byte in sequence.
1: extrw,u %r20, 7, 8, %r21
cmpclr,<> %r25, %r21, %r0
ldo -4(%r26), %r28
cmpclr,<> %r0, %r21, %r0
bv,n 0(%r2)
extrw,u %r20, 15, 8, %r21
cmpclr,<> %r25, %r21, %r0
ldo -3(%r26), %r28
cmpclr,<> %r0, %r21, %r0
bv,n 0(%r2)
extrw,u %r20, 23, 8, %r21
cmpclr,<> %r25, %r21, %r0
ldo -2(%r26), %r28
cmpclr,<> %r0, %r21, %r0
bv,n 0(%r2)
;; Having checked the others, the last byte must be NUL.
;; Do check for the unusual case of C == NUL.
cmpclr,<> %r25, %r0, %r0
ldo -1(%r26), %r28
bv,n 0(%r2)
.exit
.procend
