Estaba mirando el strlencódigo aquí y me preguntaba si las optimizaciones utilizadas en el código son realmente necesarias. Por ejemplo, ¿por qué algo como lo siguiente no funcionaría igual de bien o mejor?
unsigned long strlen(char s[]) {
unsigned long i;
for (i = 0; s[i] != '\0'; i++)
continue;
return i;
}
¿No es el código más simple mejor y / o más fácil de optimizar para el compilador?
El código de strlenen la página detrás del enlace se ve así:
/* Copyright (C) 1991, 1993, 1997, 2000, 2003 Free Software Foundation, Inc. This file is part of the GNU C Library. Written by Torbjorn Granlund (tege@sics.se), with help from Dan Sahlin (dan@sics.se); commentary by Jim Blandy (jimb@ai.mit.edu). The GNU C Library is free software; you can redistribute it and/or modify it under the terms of the GNU Lesser General Public License as published by the Free Software Foundation; either version 2.1 of the License, or (at your option) any later version. The GNU C Library is distributed in the hope that it will be useful, but WITHOUT ANY WARRANTY; without even the implied warranty of MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the GNU Lesser General Public License for more details. You should have received a copy of the GNU Lesser General Public License along with the GNU C Library; if not, write to the Free Software Foundation, Inc., 59 Temple Place, Suite 330, Boston, MA 02111-1307 USA. */ #include <string.h> #include <stdlib.h> #undef strlen /* Return the length of the null-terminated string STR. Scan for the null terminator quickly by testing four bytes at a time. */ size_t strlen (str) const char *str; { const char *char_ptr; const unsigned long int *longword_ptr; unsigned long int longword, magic_bits, himagic, lomagic; /* Handle the first few characters by reading one character at a time. Do this until CHAR_PTR is aligned on a longword boundary. */ for (char_ptr = str; ((unsigned long int) char_ptr & (sizeof (longword) - 1)) != 0; ++char_ptr) if (*char_ptr == '\0') return char_ptr - str; /* All these elucidatory comments refer to 4-byte longwords, but the theory applies equally well to 8-byte longwords. */ longword_ptr = (unsigned long int *) char_ptr; /* Bits 31, 24, 16, and 8 of this number are zero. Call these bits the "holes." Note that there is a hole just to the left of each byte, with an extra at the end: bits: 01111110 11111110 11111110 11111111 bytes: AAAAAAAA BBBBBBBB CCCCCCCC DDDDDDDD The 1-bits make sure that carries propagate to the next 0-bit. The 0-bits provide holes for carries to fall into. */ magic_bits = 0x7efefeffL; himagic = 0x80808080L; lomagic = 0x01010101L; if (sizeof (longword) > 4) { /* 64-bit version of the magic. */ /* Do the shift in two steps to avoid a warning if long has 32 bits. */ magic_bits = ((0x7efefefeL << 16) << 16) | 0xfefefeffL; himagic = ((himagic << 16) << 16) | himagic; lomagic = ((lomagic << 16) << 16) | lomagic; } if (sizeof (longword) > 8) abort (); /* Instead of the traditional loop which tests each character, 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 zero. */ for (;;) { /* We tentatively exit the loop if adding MAGIC_BITS to LONGWORD fails to change any of the hole bits of LONGWORD. 1) Is this safe? Will it catch all the zero bytes? Suppose there is a byte with all zeros. Any carry bits propagating from its left will fall into the hole at its least significant bit and stop. Since there will be no carry from its most significant bit, the LSB of the byte to the left will be unchanged, and the zero will be detected. 2) Is this worthwhile? Will it ignore everything except zero bytes? Suppose every byte of LONGWORD has a bit set somewhere. There will be a carry into bit 8. If bit 8 is set, this will carry into bit 16. If bit 8 is clear, one of bits 9-15 must be set, so there will be a carry into bit 16. Similarly, there will be a carry into bit 24. If one of bits 24-30 is set, there will be a carry into bit 31, so all of the hole bits will be changed. The one misfire occurs when bits 24-30 are clear and bit 31 is set; in this case, the hole at bit 31 is not changed. If we had access to the processor carry flag, we could close this loophole by putting the fourth hole at bit 32! So it ignores everything except 128's, when they're aligned properly. */ longword = *longword_ptr++; if ( #if 0 /* Add MAGIC_BITS to LONGWORD. */ (((longword + magic_bits) /* Set those bits that were unchanged by the addition. */ ^ ~longword) /* Look at only the hole bits. If any of the hole bits are unchanged, most likely one of the bytes was a zero. */ & ~magic_bits) #else ((longword - lomagic) & himagic) #endif != 0) { /* Which of the bytes was the zero? If none of them were, it was a misfire; continue the search. */ const char *cp = (const char *) (longword_ptr - 1); if (cp[0] == 0) return cp - str; if (cp[1] == 0) return cp - str + 1; if (cp[2] == 0) return cp - str + 2; if (cp[3] == 0) return cp - str + 3; if (sizeof (longword) > 4) { if (cp[4] == 0) return cp - str + 4; if (cp[5] == 0) return cp - str + 5; if (cp[6] == 0) return cp - str + 6; if (cp[7] == 0) return cp - str + 7; } } } } libc_hidden_builtin_def (strlen)
¿Por qué esta versión se ejecuta rápidamente?
¿No está haciendo mucho trabajo innecesario?
2
Los comentarios no son para discusión extendida; Esta conversación se ha movido al chat .
—
Samuel Liew
Para futuras referencias, el repositorio fuente oficial para GNU libc está en < sourceware.org/git/?p=glibc.git >. < sourceware.org/git/?p=glibc.git;a=blob;f=string/… > de hecho muestra un código similar al anterior; sin embargo,
—
zwol
sysdepsse usará una implementación de lenguaje ensamblador escrita a mano desde el directorio, en la mayoría de las arquitecturas compatibles de glibc (la arquitectura más utilizada que no tiene un reemplazo es MIPS).
Votar para cerrar esto como principalmente basado en opiniones; "¿Realmente se necesita xxx en xxx?" es subjetivo a las opiniones de las personas.
—
SS Anne
@ JL2210: Buen punto, arreglé el título para capturar el espíritu de la pregunta en un título que no parece que se pregunte si se necesita rendimiento, sino por qué necesitamos estas optimizaciones para obtener el rendimiento.
—
Peter Cordes
@ JL2210 FWIW, el título original era "¿Por qué es tan complejo en C [sic!]", Y se cerró como "demasiado amplio", luego se volvió a abrir y luego se cerró como "principalmente basado en la opinión". Traté de arreglar esto (entrando en el fuego cruzado de "¡rompiste mi pregunta!" Y "¡Ustedes están abusando de sus poderes de edición!" Mientras tanto), pero en mi humilde opinión el problema mintió (y aún reside) en la premisa básica de la pregunta, lo cual era problemático ("este código es demasiado complejo para que yo lo entienda" no es adecuado para preguntas y respuestas; en mi opinión, es una solicitud de tutoría, no una respuesta). No lo volveré a tocar con un poste de 60 pies :)
