strlen的实现是通过4个字节4个字节进行枚举，然后通过位运算来判断这4个字节中是否有一个字节含有0，这样的话，效率就提高了4倍。
这个效率提高是假设a&b&c&d与a&b有差不多效率的前提下。
那用8字节8字节来偏移的话，是不是更快呢？32位机上不会，64位机上会提高一倍。因为a&b在64位下会提高一倍，因为32位的寄存器大小是32位的，对于分别MOV高位与低位两次。
本来实验a&b&c&d与a&b的速度的，经实验验证，这两个效率确实是差不多的，然后去看汇编，看指令条数，在没有使用-O优化下，指令的条数差别跟运算符号的个数的倍数相同，就让我感到疑惑了。
下面附上实验的代码：
#include <iostream> #include <time.h> #include <cstdio> #include <string> using namespace std; int _strlen(const char *str) { const unsigned int *p = (const unsigned int *) str; unsigned int low = 0x01010101; unsigned int high = 0x80808080; while (true) { unsigned int d = *p++; if (((d - low) & ~d & high) != 0) { // handle [0...256) //if (((d - low) & high) != 0) { // handle [0...128) break; } } const char *q = (const char *)(p - 1); for (int i = 0; i < (int)sizeof(unsigned int); i++) { if (q[i] == 0) { return q - str + i; } } return -1; } int _strlen2(const char *str) { const char *p = str; while (*p != 0) { p++; } return p - str; } int _strlen3(const char *str) { const unsigned long long *p = (const unsigned long long *) str; unsigned long long low = 0x0101010101010101; unsigned long long high = 0x8080808080808080; while (true) { unsigned long long d = *p++; if (((d - low) & ~d & high) != 0) { // handle [0...256) //if (((d - low) & high) != 0) { // handle [0...128) break; } } const char *q = (const char *)(p - 1); for (int i = 0; i < (int)sizeof(unsigned long long); i++) { if (q[i] == 0) { return q - str + i; } } return -1; } size_t _strlen4(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 == ' ') 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; } /*j 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; } } } } string gen_data() { string a; for (int i = 0; i < 100000; i++) { a.push_back('a'); } return a; } double get_run_time(int(*fp)(const char *), const char *str, int count) { clock_t start = clock(); for (int i = 0; i < count; i++) { fp(str); } clock_t end = clock(); return (double)(end - start) / CLOCKS_PER_SEC; } double get_run_time(size_t(*fp)(const char *), const char *str, int count) { clock_t start = clock(); for (int i = 0; i < count; i++) { fp(str); } clock_t end = clock(); return (double)(end - start) / CLOCKS_PER_SEC; } int main() { string a = gen_data(); printf("%d ", _strlen(a.c_str())); printf("%d ", _strlen2(a.c_str())); printf("%d ", _strlen3(a.c_str())); printf("%d ", (int)strlen(a.c_str())); double time = get_run_time(&_strlen, a.c_str(), 10000); printf("%f ", time); double time2 = get_run_time(&_strlen2, a.c_str(), 10000); printf("%f ", time2); double time3 = get_run_time(&_strlen3, a.c_str(), 10000); printf("%f ", time3); double time4 = get_run_time(&strlen, a.c_str(), 10000); printf("%f ", time4); double time5 = get_run_time(&_strlen4, a.c_str(), 10000); printf("%f ", time5); }