9412b23450
The generic implementation of strlen() reads strings byte per byte. This patch implements strlen() in assembly based on a read of entire words, in the same spirit as what some other arches and glibc do. On a 8xx the time spent in strlen is reduced by 3/4 for long strings. strlen() selftest on an 8xx provides the following values: Before the patch (ie with the generic strlen() in lib/string.c): len 256 : time = 1.195055 len 016 : time = 0.083745 len 008 : time = 0.046828 len 004 : time = 0.028390 After the patch: len 256 : time = 0.272185 ==> 78% improvment len 016 : time = 0.040632 ==> 51% improvment len 008 : time = 0.033060 ==> 29% improvment len 004 : time = 0.029149 ==> 2% degradation On a 832x: Before the patch: len 256 : time = 0.236125 len 016 : time = 0.018136 len 008 : time = 0.011000 len 004 : time = 0.007229 After the patch: len 256 : time = 0.094950 ==> 60% improvment len 016 : time = 0.013357 ==> 26% improvment len 008 : time = 0.010586 ==> 4% improvment len 004 : time = 0.008784 Signed-off-by: Christophe Leroy <christophe.leroy@c-s.fr> Signed-off-by: Michael Ellerman <mpe@ellerman.id.au>
78 lines
2.6 KiB
ArmAsm
78 lines
2.6 KiB
ArmAsm
/* SPDX-License-Identifier: GPL-2.0 */
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/*
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* strlen() for PPC32
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*
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* Copyright (C) 2018 Christophe Leroy CS Systemes d'Information.
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*
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* Inspired from glibc implementation
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*/
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#include <asm/ppc_asm.h>
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#include <asm/export.h>
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#include <asm/cache.h>
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.text
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/*
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* Algorithm:
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*
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* 1) Given a word 'x', we can test to see if it contains any 0 bytes
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* by subtracting 0x01010101, and seeing if any of the high bits of each
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* byte changed from 0 to 1. This works because the least significant
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* 0 byte must have had no incoming carry (otherwise it's not the least
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* significant), so it is 0x00 - 0x01 == 0xff. For all other
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* byte values, either they have the high bit set initially, or when
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* 1 is subtracted you get a value in the range 0x00-0x7f, none of which
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* have their high bit set. The expression here is
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* (x - 0x01010101) & ~x & 0x80808080), which gives 0x00000000 when
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* there were no 0x00 bytes in the word. You get 0x80 in bytes that
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* match, but possibly false 0x80 matches in the next more significant
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* byte to a true match due to carries. For little-endian this is
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* of no consequence since the least significant match is the one
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* we're interested in, but big-endian needs method 2 to find which
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* byte matches.
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* 2) Given a word 'x', we can test to see _which_ byte was zero by
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* calculating ~(((x & ~0x80808080) - 0x80808080 - 1) | x | ~0x80808080).
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* This produces 0x80 in each byte that was zero, and 0x00 in all
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* the other bytes. The '| ~0x80808080' clears the low 7 bits in each
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* byte, and the '| x' part ensures that bytes with the high bit set
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* produce 0x00. The addition will carry into the high bit of each byte
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* iff that byte had one of its low 7 bits set. We can then just see
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* which was the most significant bit set and divide by 8 to find how
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* many to add to the index.
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* This is from the book 'The PowerPC Compiler Writer's Guide',
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* by Steve Hoxey, Faraydon Karim, Bill Hay and Hank Warren.
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*/
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_GLOBAL(strlen)
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andi. r0, r3, 3
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lis r7, 0x0101
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addi r10, r3, -4
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addic r7, r7, 0x0101 /* r7 = 0x01010101 (lomagic) & clear XER[CA] */
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rotlwi r6, r7, 31 /* r6 = 0x80808080 (himagic) */
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bne- 3f
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.balign IFETCH_ALIGN_BYTES
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1: lwzu r9, 4(r10)
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2: subf r8, r7, r9
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and. r8, r8, r6
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beq+ 1b
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andc. r8, r8, r9
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beq+ 1b
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andc r8, r9, r6
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orc r9, r9, r6
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subfe r8, r6, r8
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nor r8, r8, r9
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cntlzw r8, r8
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subf r3, r3, r10
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srwi r8, r8, 3
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add r3, r3, r8
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blr
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/* Missaligned string: make sure bytes before string are seen not 0 */
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3: xor r10, r10, r0
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orc r8, r8, r8
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lwzu r9, 4(r10)
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slwi r0, r0, 3
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srw r8, r8, r0
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orc r9, r9, r8
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b 2b
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EXPORT_SYMBOL(strlen)
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