From patchwork Sat Dec 17 06:57:21 2016 Content-Type: text/plain; charset="utf-8" MIME-Version: 1.0 Content-Transfer-Encoding: 7bit X-Patchwork-Submitter: Richard Henderson X-Patchwork-Id: 706649 Return-Path: X-Original-To: incoming@patchwork.ozlabs.org Delivered-To: patchwork-incoming@bilbo.ozlabs.org Received: from sourceware.org (server1.sourceware.org [209.132.180.131]) (using TLSv1.2 with cipher ECDHE-RSA-AES256-GCM-SHA384 (256/256 bits)) (No client certificate requested) by ozlabs.org (Postfix) with ESMTPS id 3tgdL961Ymz9t2D for ; Sat, 17 Dec 2016 17:58:09 +1100 (AEDT) Authentication-Results: ozlabs.org; dkim=pass (1024-bit key; secure) header.d=sourceware.org header.i=@sourceware.org header.b="cwgK/TWY"; dkim-atps=neutral DomainKey-Signature: a=rsa-sha1; c=nofws; d=sourceware.org; h=list-id :list-unsubscribe:list-subscribe:list-archive:list-post :list-help:sender:from:to:subject:date:message-id:in-reply-to :references; q=dns; s=default; b=DXOLwgtJc8qr4PdUmWUBF8yTGndJ3lK Tz0Vh8vJOoSUBQmgy/pBOLWYCnsx1oTTL1bp49mhRgTzoF7XsKmK4eK8MyJCWuj+ lSrvsDa5t9UCk8F0lVYB41O5g6d/OBobLYYPmsRolH+dFbr6gIbhp8sbYn11UIXA k/x8mZGS61cc= DKIM-Signature: v=1; a=rsa-sha1; c=relaxed; d=sourceware.org; h=list-id :list-unsubscribe:list-subscribe:list-archive:list-post :list-help:sender:from:to:subject:date:message-id:in-reply-to :references; s=default; bh=Y7sbUXmz4Irtaslpsjx5F6PhkcU=; b=cwgK/ TWYn5bWmWX2IRxhpS+aRj7Biacpki/cGtlXtf/c68ocrJW4ps9DhCC5mZfi17GvG iO2wRkDg5Okjy0wkmS+AVu/UzJ7qEIlAftXP5RhTS/wRpRd/BShirhqWla5qC9pp QZnR8XgfLzlLZVag2dcqrRuJBZsIrmthjYsIwo= Received: (qmail 92288 invoked by alias); 17 Dec 2016 06:57:45 -0000 Mailing-List: contact libc-alpha-help@sourceware.org; run by ezmlm Precedence: bulk List-Id: List-Unsubscribe: List-Subscribe: List-Archive: List-Post: List-Help: , Sender: libc-alpha-owner@sourceware.org Delivered-To: mailing list libc-alpha@sourceware.org Received: (qmail 92188 invoked by uid 89); 17 Dec 2016 06:57:44 -0000 Authentication-Results: sourceware.org; auth=none X-Virus-Found: No X-Spam-SWARE-Status: No, score=-1.6 required=5.0 tests=BAYES_00, FREEMAIL_ENVFROM_END_DIGIT, FREEMAIL_FROM, RCVD_IN_DNSWL_NONE, SPF_PASS autolearn=no version=3.3.2 spammy=j1, uintmax_t, tends, HX-Received:10.99.112.66 X-HELO: mail-pg0-f67.google.com X-Google-DKIM-Signature: v=1; a=rsa-sha256; c=relaxed/relaxed; d=1e100.net; s=20161025; h=x-gm-message-state:sender:from:to:subject:date:message-id :in-reply-to:references; bh=QrzFZiQMTipidSSwtrWdKa7d30ocBj2zJmmMry6YxVo=; b=HYJauEWKzelRl61OtzrR/r43SD2ERPMEIAh3BKYxLCdEv887ZumniWuTVv3/p/1iGx Rk+jcKlj7bfKBc37IY8seepqVxjHRwKfyNo3gIO1M7YsDeTse5Hx+w48HLbJKrZ8VNEL kHhG2DSJ4UI3TGV1++WRp219JqH1n367+g04t7rSsH4bJanm9JWt1KggyMfnv0nIjEI9 H5ScH+oCbCpbrd0LPP8TbsEjWtWK15e5aYL1SxaCVggUpimeYs4Sfh57fTS7NAy+5dHT DqQES46JWnQjCs/dPpXK8JZKakCSTBkXO1WO8tHF/RkHz+02ajIXL2ZFmzseA5H1Qs6M jBDA== X-Gm-Message-State: AKaTC00kSe2psmBKpeJyKTqNY7k/IjUIYnV/pT5qUl/p/oANziSxYqISFRKIutBQeayuNA== X-Received: by 10.99.112.66 with SMTP id a2mr12412060pgn.43.1481957860955; Fri, 16 Dec 2016 22:57:40 -0800 (PST) From: Richard Henderson To: libc-alpha@sourceware.org Subject: [PATCH 03/11] Improve generic memchr Date: Fri, 16 Dec 2016 22:57:21 -0800 Message-Id: <20161217065729.28561-4-rth@twiddle.net> In-Reply-To: <20161217065729.28561-1-rth@twiddle.net> References: <20161217065729.28561-1-rth@twiddle.net> * string/memchr.c: Use haszero.h, whichzero.h. --- string/memchr.c | 118 ++++++++++++++------------------------------------------ 1 file changed, 29 insertions(+), 89 deletions(-) diff --git a/string/memchr.c b/string/memchr.c index ca9fd99..858a494 100644 --- a/string/memchr.c +++ b/string/memchr.c @@ -25,10 +25,11 @@ #endif #include - #include - #include +#include +#include +#include #undef __memchr #ifdef _LIBC @@ -47,111 +48,50 @@ void * MEMCHR (void const *s, int c_in, size_t n) { - /* On 32-bit hardware, choosing longword to be a 32-bit unsigned - long instead of a 64-bit uintmax_t tends to give better - performance. On 64-bit hardware, unsigned long is generally 64 - bits already. Change this typedef to experiment with - performance. */ - typedef unsigned long int longword; - const unsigned char *char_ptr; - const longword *longword_ptr; - longword repeated_one; - longword repeated_c; + const unsigned long int *longword_ptr; + unsigned long int longword, repeated_c; unsigned char c; + uintptr_t i, align; c = (unsigned char) c_in; + char_ptr = (const unsigned char *) s; /* Handle the first few bytes by reading one byte at a time. Do this until CHAR_PTR is aligned on a longword boundary. */ - for (char_ptr = (const unsigned char *) s; - n > 0 && (size_t) char_ptr % sizeof (longword) != 0; - --n, ++char_ptr) + align = -(uintptr_t)char_ptr % sizeof(longword); + if (align > n) + align = n; + for (i = 0; i < align; ++i, ++char_ptr) if (*char_ptr == c) return (void *) char_ptr; - longword_ptr = (const longword *) char_ptr; + /* Set up a longword, each of whose bytes is C. */ + repeated_c = (-1ul / 0xff) * c; - /* All these elucidatory comments refer to 4-byte longwords, - but the theory applies equally well to any size longwords. */ + longword_ptr = (const unsigned long int *) char_ptr; + n -= align; + if (__glibc_unlikely(n == 0)) + return NULL; - /* Compute auxiliary longword values: - repeated_one is a value which has a 1 in every byte. - repeated_c has c in every byte. */ - repeated_one = 0x01010101; - repeated_c = c | (c << 8); - repeated_c |= repeated_c << 16; - if (0xffffffffU < (longword) -1) + /* Loop while we have more than one longword remaining. */ + while (n > sizeof (longword)) { - repeated_one |= repeated_one << 31 << 1; - repeated_c |= repeated_c << 31 << 1; - if (8 < sizeof (longword)) - { - size_t i; - - for (i = 64; i < sizeof (longword) * 8; i *= 2) - { - repeated_one |= repeated_one << i; - repeated_c |= repeated_c << i; - } - } - } - - /* Instead of the traditional loop which tests each byte, we will test a - longword at a time. The tricky part is testing if *any of the four* - bytes in the longword in question are equal to c. We first use an xor - with repeated_c. This reduces the task to testing whether *any of the - four* bytes in longword1 is zero. - - We compute tmp = - ((longword1 - repeated_one) & ~longword1) & (repeated_one << 7). - That is, we perform the following operations: - 1. Subtract repeated_one. - 2. & ~longword1. - 3. & a mask consisting of 0x80 in every byte. - Consider what happens in each byte: - - If a byte of longword1 is zero, step 1 and 2 transform it into 0xff, - and step 3 transforms it into 0x80. A carry can also be propagated - to more significant bytes. - - If a byte of longword1 is nonzero, let its lowest 1 bit be at - position k (0 <= k <= 7); so the lowest k bits are 0. After step 1, - the byte ends in a single bit of value 0 and k bits of value 1. - After step 2, the result is just k bits of value 1: 2^k - 1. After - step 3, the result is 0. And no carry is produced. - So, if longword1 has only non-zero bytes, tmp is zero. - Whereas if longword1 has a zero byte, call j the position of the least - significant zero byte. Then the result has a zero at positions 0, ..., - j-1 and a 0x80 at position j. We cannot predict the result at the more - significant bytes (positions j+1..3), but it does not matter since we - already have a non-zero bit at position 8*j+7. - - So, the test whether any byte in longword1 is zero is equivalent to - testing whether tmp is nonzero. */ - - while (n >= sizeof (longword)) - { - longword longword1 = *longword_ptr ^ repeated_c; - - if ((((longword1 - repeated_one) & ~longword1) - & (repeated_one << 7)) != 0) - break; + longword = *longword_ptr ^ repeated_c; + if (haszero(longword)) + goto found; longword_ptr++; n -= sizeof (longword); } - char_ptr = (const unsigned char *) longword_ptr; - - /* At this point, we know that either n < sizeof (longword), or one of the - sizeof (longword) bytes starting at char_ptr is == c. On little-endian - machines, we could determine the first such byte without any further - memory accesses, just by looking at the tmp result from the last loop - iteration. But this does not work on big-endian machines. Choose code - that works in both cases. */ - - for (; n > 0; --n, ++char_ptr) + /* Since our pointer is aligned, we can always read the last longword. */ + longword = *longword_ptr ^ repeated_c; + if (haszero(longword)) { - if (*char_ptr == c) - return (void *) char_ptr; + found: + i = whichzero(longword); + if (i < n) + return (char *) longword_ptr + i; } return NULL;