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分享阅读gcc libc的源代码wcslen函数的疑问
为什么连续用了3个同样的if语句？真是不解。
源代码如下：
版本信息：
glibc 2.0
源代码如下：
size_t
wcslen (s)
const wchar_t *s;
{
size_t len = 0;
while (s[len] != L'\0')
{
if (s[++len] == L'\0')
return len;
if (s[++len] == L'\0')
return len;
if (s[++len] == L'\0')
return len;
++len;
}
return len;
}
版本信息：
glibc 2.0
...全文
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fuabcck 2015-11-30
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执行4次是为了字节对齐
z2357 2014-03-07
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看到了那个多字节吗？是多字节的原因。
mLee79 2013-05-07
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gcc的libc，大概有这样的假定：编译我代码的编译器，可能是“土包子”，可能相当老旧。看libc的代码语法，也可看出，很土，C语言的一点点的更新，都不敢用。
[/quote]
glibc是要用首次静态编译的gcc编译的, 这时候很多语言特性本身就不能使用, 我们使用的gcc是要在编译完glibc以后, 使用新鲜出炉的glibc库再次编译gcc的代码生成的, 这时候才能支持所有的语言特性, 像glibc这种基础库能够使用的语言特性是要受到限制的..
john 2013-05-06
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关于这段代码，如此的计较优化质量，我的看法是，如果要编写系统函数，如本程序，的确的需要如此斤斤计较，尽力的优化。但如果编写一般的用户程序，就一切委托给编译器了，编写程序的专注点，不应该在于协助编译器优化代码上了。在学习语言阶段，可以锱铢必较，甚至研究一下汇编结果，但......
john 2013-05-06
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gcc的libc，大概有这样的假定：编译我代码的编译器，可能是“土包子”，可能相当老旧。看libc的代码语法，也可看出，很土，C语言的一点点的更新，都不敢用。
日立奔腾浪潮微软松下联想 2013-05-03
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就是循环展开优化，减少分支指令执行的次数，也就减少了处理器分支预测可能失败的次数。
相当老的优化技巧了，90年代的watcom c++编译器最爱使用这种套路，对于较新的处理器，这种优化效果不明显。
日立奔腾浪潮微软松下联想 2013-05-03
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哎呀，15楼贴错了，请版主删除。
日立奔腾浪潮微软松下联想 2013-05-03
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相对于程序加载的起始地址是固定的，但是不能保证每次加载/在不同的系统上加载时，静态变量的地址(逻辑地址)都固定，因为操作系统的加载过程可能对其重定位到不同的起始地址。
至于其在物理内存中的地址，运行时都有可能改变。
john 2013-05-02
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顺这你的思路，启发一个思路：这样同样的语句重叠放置，也许是为了利用CPU的流水线功能。现在ARM、DSP也都有这个设计。一条指令执行，如果需要4个周期，4个同样的指令连续执行，需要周期数不是4*4 = 16 ，而可能是4 + 3*1 = 7。
john 2013-05-02
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流水线；
循环展开优化 ；
如果没有优化也没有害处。
就是这些。
CyberLogix 2013-05-02
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循环展开优化 啊
mLee79 2013-05-02
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我看了下 x86 , x64 , arm , ppc 下的目标代码, 这样写与平凡的写法生成的目标代码基本没有啥区别 (-O3 -fomit-frame-pointer) , 看来gcc的优化还没有足够强大, 也许只是写的人习惯了看到循环就做循环展开优化, 反正也不会变的更坏, 本来我期望 arm 下能生成类似的指令的 ...
ldr ..
cmp ..
beq ..
ldrne ..
cmpne ..
beq ..
john 2013-04-27
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是一个思路，待有时间一定看一下，编译处理的机器码。估计应该将各种优化均设置上，才能看出端倪。
AndyStevens 2013-04-27
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不准确的猜想：可能是为了优化代码执行效率，c循环汇编之后，由于编译器的缘故，大多效率并不高，减少循环复杂度是提高效率的一个可行的方法
john 2013-04-27
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另一个版本glibc 2.7 wcslen.c
看样子是有意为止。
* Copyright (C) 1995, 1996, 1997, 1998, 2011 Free Software Foundation, Inc.
This file is part of the GNU C Library.
Contributed by Ulrich Drepper <drepper@gnu.ai.mit.edu>, 1995.
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, see
<http://www.gnu.org/licenses/>. */
#include <wchar.h>
/* Return length of string S. */
#ifdef WCSLEN
# define __wcslen WCSLEN
#endif
size_t
__wcslen (s)
const wchar_t *s;
{
size_t len = 0;
while (s[len] != L'\0')
{
if (s[++len] == L'\0')
return len;
if (s[++len] == L'\0')
return len;
if (s[++len] == L'\0')
return len;
++len;
}
return len;
}
#ifndef WCSLEN
weak_alias (__wcslen, wcslen)
#endif
/* Copyright (C) 1991,1993,1997,2000,2003,2009 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, see
<http://www.gnu.org/licenses/>. */
#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, 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. */
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. */
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 (;;)
{
longword = *longword_ptr++;
if (((longword - lomagic) & ~longword & himagic) != 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)
赵4老师 2013-04-27
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对学习编程者的忠告：
眼过千遍不如手过一遍！
书看千行不如手敲一行！
手敲千行不如单步一行！
单步源代码千行不如单步对应汇编一行！
赵4老师 2013-04-27
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先
http://www.microsoft.com/visualstudio/chs/downloads#d-2010-express
点开Visual C++ 2010 Express下面的语言选‘简体中文’，再点立即安装
再参考C:\Program Files\Microsoft Visual Studio 10.0\VC\crt\src\intel\strlen.asm：
page ,132
title strlen - return the length of a null-terminated string
;***
;strlen.asm - contains strlen() routine
;
; Copyright (c) Microsoft Corporation. All rights reserved.
;
;Purpose:
; strlen returns the length of a null-terminated string,
; not including the null byte itself.
;
;*******************************************************************************
.xlist
include cruntime.inc
.list
page
;***
;strlen - return the length of a null-terminated string
;
;Purpose:
; Finds the length in bytes of the given string, not including
; the final null character.
;
; Algorithm:
; int strlen (const char * str)
; {
; int length = 0;
;
; while( *str++ )
; ++length;
;
; return( length );
; }
;
;Entry:
; const char * str - string whose length is to be computed
;
;Exit:
; EAX = length of the string "str", exclusive of the final null byte
;
;Uses:
; EAX, ECX, EDX
;
;Exceptions:
;
;*******************************************************************************
CODESEG
public strlen
strlen proc \
buf:ptr byte
OPTION PROLOGUE:NONE, EPILOGUE:NONE
.FPO ( 0, 1, 0, 0, 0, 0 )
string equ [esp + 4]
mov ecx,string ; ecx -> string
test ecx,3 ; test if string is aligned on 32 bits
je short main_loop
str_misaligned:
; simple byte loop until string is aligned
mov al,byte ptr [ecx]
add ecx,1
test al,al
je short byte_3
test ecx,3
jne short str_misaligned
add eax,dword ptr 0 ; 5 byte nop to align label below
align 16 ; should be redundant
main_loop:
mov eax,dword ptr [ecx] ; read 4 bytes
mov edx,7efefeffh
add edx,eax
xor eax,-1
xor eax,edx
add ecx,4
test eax,81010100h
je short main_loop
; found zero byte in the loop
mov eax,[ecx - 4]
test al,al ; is it byte 0
je short byte_0
test ah,ah ; is it byte 1
je short byte_1
test eax,00ff0000h ; is it byte 2
je short byte_2
test eax,0ff000000h ; is it byte 3
je short byte_3
jmp short main_loop ; taken if bits 24-30 are clear and bit
; 31 is set
byte_3:
lea eax,[ecx - 1]
mov ecx,string
sub eax,ecx
ret
byte_2:
lea eax,[ecx - 2]
mov ecx,string
sub eax,ecx
ret
byte_1:
lea eax,[ecx - 3]
mov ecx,string
sub eax,ecx
ret
byte_0:
lea eax,[ecx - 4]
mov ecx,string
sub eax,ecx
ret
strlen endp
end
ForestDB 2013-04-26
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不准确的猜想：每次使得len加4。
john 2013-04-26
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哦。这个并行处理的知识还真是不知道。谢谢提示。有时间看看有关的资料。
flyound 2013-04-26
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这个应该是对CPU指令执行的优化吧，并行处理，如果每次都进入while就不会进行并行运算了。
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