一般來說,一個入門的程式員大概可以不費勁的寫出這樣的 strlen 代碼。
size_t my_strlen(const char * str) {
<span style="white-space:pre"> </span>size_t length = 0 ;
<span style="white-space:pre"> </span>while (*str++ )
<span style="white-space:pre"> </span>++ length;
<span style="white-space:pre"> </span>return length;
}
那我們再來看看,glibc函數庫裡面的strlen是如何實作的。
#include <string.h>
#include <stdlib.h>
<pre name="code" class="cpp">size_t strlen(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;
}
}
}
}
很明顯這段代碼比我們自己寫的版本要複雜多了。其中注釋寫的比較清楚,我也不翻譯了。
下面主要對其中幾個點做分析吧。如果要看更詳細的,請點選《strlen源碼剖析》
簡單版本和glibc版本主要的差別在于:通過記憶體對齊,來加快CPU的讀取速度。
(在計算機在讀取記憶體中的資料的時候,在記憶體對齊的狀态下一次讀取一個word的資料,是最節省時間的。例如在32位的計算機中,一個WORD為4 byte,則WORD資料的起始位址能被4整除的時候CPU的存取效率比較高。)
其中用到了兩個技巧:
(1)由于傳進來的字元串的位址有可能不是4位元組(long int)對其的,是以首先找到4位元組對其的那個位址
(2)技巧就是如何高效的判斷在讀取的4個位元組中是否有位元組為0
是以這個算法的整體思路如下:
(1)從字元串的開頭開始,一次判斷一個字元直到記憶體對齊(也就是位址能被4整除)。
(2)如果在記憶體對齊之前,字元串就已經結束了(也就是讀取到'\0'),那麼就return,否則到繼續下一步;
(3)一次讀入并判斷一個DWORD,如果此DWORD中沒有為0的位元組,則繼續下一個DWORD;如果判斷出有'\0',就跳到下一步。
(4)找出該DWORD中第一個為0的位元組的位置,然後return。
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