glibc中strlen的实现
glibc中的strlen的实现主要的思想就是每次检测4个字节(long int)。这样的话就降低了循环的次数，从而从整体上提高了效率。这里它使用了两个技巧，一个是由于传进来的字符串的地址有可能不是4字节(long int)对其的，因此首先需要遍历字符串从而找到4字节对其的那个地址。然后再进行比较.
第二个技巧就是如何高效的判断4个字节中是否有字节为0.
下来来看源码，这个源码的注释还是满详细的。这里主要都是一些位计算的技巧：
size_tstrlen (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) return cp - str;if (cp == 0) return cp - str + 1;if (cp == 0) return cp - str + 2;if (cp == 0) return cp - str + 3;if (sizeof (longword) > 4) { if (cp == 0)return cp - str + 4; if (cp == 0)return cp - str + 5; if (cp == 0)return cp - str + 6; if (cp == 0)return cp - str + 7; }} }}
代码比较长,不过注释也比较详细,这里主要分析下两个位运算:
for (char_ptr = str; ((unsigned long int) char_ptr& (sizeof (longword) - 1)) != 0; ++char_ptr) if (*char_ptr == '\0') return char_ptr - str;
这里是做一个循环来判断4-byte对齐的那个地址,这里一般的编译器传递进来的字符串指针都是4-byte对其的,因此这步在很多编译器都是跳过的.
下来来看如何在4字节的数字中判断是否有某个字节为0.
himagic = 0x80808080L;lomagic = 0x01010101L;if (((longword - lomagic) & ~longword & himagic) != 0)
这里分两个部分看.
第一部分(longword - lomagic) 这个表达式是用来判断最高位的,它的意思是: 如果longword的任何一个字节如果大于0x80或者等于0的情况下,它的这个字节的最高位的值都会是不等于0的.
第二部分(~longword & himagic)这个表达式是用来判断longword的每个字节的最高位的设置,也就是判断longword是否小于0x80,如果小于0x80则这个表达式的值是himagic(0x80808080).否则就是0.
因此这两部分合并起来刚好就是如果longword的某个字节小于0x80,并且某个字节为0,则整个表达式的值不为0.
而我们知道传进来的字符串的每个字符的大小的ascii码的范围就是,也就是肯定是小于0x80的.所以这个表达式就刚好是我们所要的.
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