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214c203c00
A new fgets implementation saves about 25-50% of the time spent parsing the logs. Percentile calculation has been added for timers using -pct.
148 lines
3.5 KiB
C
148 lines
3.5 KiB
C
/*
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* fast fgets() replacement for log parsing
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*
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* Copyright 2000-2009 Willy Tarreau <w@1wt.eu>
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*
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* This program is free software; you can redistribute it and/or
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* modify it under the terms of the GNU General Public License
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* as published by the Free Software Foundation; either version
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* 2 of the License, or (at your option) any later version.
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*
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* This function manages its own buffer and returns a pointer to that buffer
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* in order to avoid expensive memory copies. It also checks for line breaks
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* 32 bits at a time. It could be improved a lot using mmap() but we would
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* not be allowed to replace trailing \n with zeroes and we would be limited
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* to small log files on 32-bit machines.
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*
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*/
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#include <stdlib.h>
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#include <string.h>
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#include <stdio.h>
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#include <unistd.h>
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// return 1 if the integer contains at least one zero byte
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static inline unsigned int has_zero(unsigned int x)
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{
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if (!(x & 0xFF000000U) ||
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!(x & 0xFF0000U) ||
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!(x & 0xFF00U) ||
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!(x & 0xFFU))
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return 1;
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return 0;
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}
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static inline unsigned int has_zero64(unsigned long long x)
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{
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unsigned long long x2;
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x2 = x & (x >> 8);
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/* no need to split it further */
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if ((x2 & 0x00FF) && (x2 & 0x00FF0000) && (x2 & 0x00FF00000000ULL) && (x2 & 0x00FF000000000000ULL))
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return 0; // approx 11/12 return here
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if (!(x & 0xff00000000000000ULL) ||
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!(x & 0xff000000000000ULL) ||
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!(x & 0xff0000000000ULL) ||
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!(x & 0xff00000000ULL) ||
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!(x & 0xff000000UL) ||
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!(x & 0xff0000UL) ||
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!(x & 0xff00UL) ||
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!(x & 0xffUL))
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return 1; // approx 1/3 of the remaining return here
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return 0;
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}
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#define FGETS2_BUFSIZE (256*1024)
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const char *fgets2(FILE *stream)
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{
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static char buffer[FGETS2_BUFSIZE + 5];
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static char *end = buffer;
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static char *line = buffer;
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char *next;
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int ret;
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next = line;
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while (1) {
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/* this is a speed-up, we read 32 bits at once and check for an
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* LF character there. We stop if found then continue one at a
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* time.
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*/
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while (next < end && (((unsigned long)next) & 7) && *next != '\n')
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next++;
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/* now next is multiple of 4 or equal to end */
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while (next <= (end-32)) {
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if (has_zero64(*(unsigned long long *)next ^ 0x0A0A0A0A0A0A0A0AULL))
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break;
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next += 8;
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if (has_zero64(*(unsigned long long *)next ^ 0x0A0A0A0A0A0A0A0AULL))
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break;
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next += 8;
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if (has_zero64(*(unsigned long long *)next ^ 0x0A0A0A0A0A0A0A0AULL))
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break;
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next += 8;
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if (has_zero64(*(unsigned long long *)next ^ 0x0A0A0A0A0A0A0A0AULL))
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break;
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next += 8;
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}
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/* we finish if needed. Note that next might be slightly higher
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* than end here because we might have gone past it above.
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*/
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while (next < end) {
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if (*next == '\n') {
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const char *start = line;
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*next = '\0';
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line = next + 1;
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return start;
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}
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next++;
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}
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/* we found an incomplete line. First, let's move the
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* remaining part of the buffer to the beginning, then
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* try to complete the buffer with a new read.
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*/
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if (line > buffer) {
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if (end != line)
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memmove(buffer, line, end - line);
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end = buffer + (end - line);
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next = end;
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line = buffer;
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} else {
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if (end == buffer + FGETS2_BUFSIZE)
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return NULL;
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}
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ret = read(fileno(stream), end, buffer + FGETS2_BUFSIZE - end);
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if (ret <= 0) {
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if (end == line)
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return NULL;
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*end = '\0';
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return line;
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}
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end += ret;
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/* search for '\n' again */
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}
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}
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#ifdef BENCHMARK
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int main() {
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const char *p;
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unsigned int lines = 0;
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while ((p=fgets2(stdin)))
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lines++;
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printf("lines=%d\n", lines);
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return 0;
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}
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#endif
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