glibc系列之strlen（）函数学习。

首先奉上的是strlen的源代码：

size_t STRLEN (const char *str) 可以看到返回值是个size_t，输入参数是char型数组。 { 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; 这里在做的是字节对齐。 ++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; 这两个魔数很重要。 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) 主要是判断这连续的四个字节有没有0x0，如果有0，那么进入到if语句中， { /* 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; } } } }

下面是代码解析

首先给出俩例子来形象的说明这个问题。 1 himagic = 0x80808080L; 1000-0000 1000-0000 1000-0000 1000-0000 lomagic = 0x01010101L; 0000-0001 0000-0001 0000-0001 0000-0001 ((longword - lomagic) & ~longword & himagic) != 0 举个例子来说明这个问题： longword = 000000ff a =(longword - lomagic) =fefefffe b = ~longword = ffffff00 a&b = fefeff00 himagic = 80808080 a&b&himagic 80808000

2 另一个例子： longword =00000000 a =(longword - lomagic) =fefefeff

b = ~longword =ffffffff

a&b = fefefeff

himagic = 80808080

a&b&himagic = 80808080

其实最主要的是if (((longword - lomagic) & ~longword & himagic) != 0) 怎么理解的。

首先考虑到的是ASCII码最高位是是0， ~longword & himagic运算的结果就是看看哪个字节的最高位不是0.如果都是0 ，则运算结果是0x80808080. longword - lomagic 运算结果是一旦longword 有一个字节为0，那么久会是0xfe。也就是找到了‘\0’,进入了if体里。

下面是我自己写的strlen（）函数。

#define ulint unsigned long int //宏定义ulint unsigned int mglStrLen(const char * str) { //allign const char *tempStr ; for (tempStr = str; (ulint)tempStr &(sizeof(ulint) - 1) != 0; tempStr++) { if (*tempStr == '\0') return tempStr - str; } //caculate the length ulint himagic = 0x80808080; ulint lomagic = 0x01010101; const ulint* longwordPtr = ( const ulint *)tempStr; //这里也要注意对const char* 强制类型转化成const ulint*时也需要( const ulint *) 。这里的const修饰很有意思。 while (1) { longwordPtr++; if (((*longwordPtr - lomagic)&~(*longwordPtr)&himagic) != 0) //跟glibc一样的处理方法。 { if (*(char *)longwordPtr == 0) //这里要注意的是longwordPtr 指向的int型，对于单个的字节比对，需要（char*）强制类型转换。 return (char *)longwordPtr - str; //这里的强制类型转化是因为str是const char* ，他们要一样。 if (*((char *)longwordPtr+1) == 0) return (char *)longwordPtr - str+1; if (*((char *)longwordPtr + 2) == 0) return (char *)longwordPtr - str + 2; if (*((char *)longwordPtr + 3) == 0) return (char *)longwordPtr - str + 3; } } } void main() { char str[] = "mglhahaahhahah"; ---->14个字节 int len = mglStrLen(str); cout << "mgl " <<len<< endl; }

程序运行结果：

mgl 14 请按任意键继续. . .

一个属于mgl的完美unsigned int strlen（const char * str）{}函数。

哈哈！！