Saya melihat-lihat strlenkode di sini dan saya bertanya-tanya apakah optimasi yang digunakan dalam kode benar-benar diperlukan? Misalnya, mengapa hal seperti ini tidak akan berfungsi sama baiknya atau lebih baik?
unsigned long strlen(char s[]) {
unsigned long i;
for (i = 0; s[i] != '\0'; i++)
continue;
return i;
}Bukankah kode sederhana lebih baik dan / atau lebih mudah untuk dioptimalkan oleh kompiler?
Kode strlenpada halaman di belakang tautan terlihat seperti ini:
/* 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)
Mengapa versi ini berjalan cepat?
Bukankah itu melakukan banyak pekerjaan yang tidak perlu?
2
Komentar bukan untuk diskusi panjang; percakapan ini telah dipindahkan ke obrolan .
—
Samuel Liew
Untuk referensi di masa mendatang, repositori sumber resmi untuk libc GNU ada di < sourceware.org/git/?p=glibc.git >. < sourceware.org/git/?p=glibc.git;a=blob;f=string/… > memang menampilkan kode yang mirip dengan di atas; namun, implementasi bahasa rakitan tulisan tangan dari
—
zwol
sysdepsdirektori akan digunakan sebagai gantinya, pada sebagian besar arsitektur yang didukung glibc (arsitektur yang paling umum digunakan yang tidak memiliki penggantian adalah MIPS).
Voting untuk menutup ini terutama didasarkan pada opini; "Apakah xxx benar-benar dibutuhkan dalam xxx?" subjektif untuk pendapat orang.
—
SS Anne
@ JL2210: Poin bagus, perbaiki judul untuk menangkap semangat pertanyaan dalam judul yang tidak terdengar seperti bertanya-tanya apakah kinerja diperlukan, hanya mengapa kita perlu optimasi ini untuk mendapatkan kinerja.
—
Peter Cordes
@ JL2210 FWIW, judul aslinya adalah "Mengapa strlen sangat kompleks di C [sic!]", Dan ditutup sebagai "terlalu luas", kemudian dibuka kembali, lalu ditutup sebagai "terutama berdasarkan pendapat". Saya mencoba untuk memperbaiki ini (mendapatkan di baku tembak "Anda memecahkan pertanyaan saya!" Dan "kalian menyalahgunakan kekuatan editing Anda!" Sementara itu), tetapi IMVHO masalahnya terletak (dan masih terletak) dalam premis dasar pertanyaan, yang bermasalah ("kode ini terlalu rumit untuk saya pahami" tidak cocok untuk T&J - IMO itu adalah permintaan untuk les, bukan untuk jawaban). Saya tidak menyentuhnya lagi dengan tiang 60 kaki :)
