關於strlen

strlen的實現是經過4個字節4個字節進行枚舉，而後經過位運算來判斷這4個字節中是否有一個字節含有0，這樣的話，效率就提升了4倍。

這個效率提升是假設a&b&c&d與a&b有差很少效率的前提下。

那用8字節8字節來偏移的話，是否是更快呢？32位機上不會，64位機上會提升一倍。由於a&b在64位下會提升一倍，由於32位的寄存器大小是32位的，對於分別MOV高位與低位兩次。

原本實驗a&b&c&d與a&b的速度的，經實驗驗證，這兩個效率確實是差很少的，而後去看彙編，看指令條數，在沒有使用-O優化下，指令的條數差異跟運算符號的個數的倍數相同，就讓我感到疑惑了。

下面附上實驗的代碼：

#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 == '\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;
    }
    /*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\n", _strlen(a.c_str()));
    printf("%d\n", _strlen2(a.c_str()));
    printf("%d\n", _strlen3(a.c_str()));
    printf("%d\n", (int)strlen(a.c_str()));
    double time = get_run_time(&_strlen, a.c_str(), 10000);
    printf("%f\n", time);
    double time2 = get_run_time(&_strlen2, a.c_str(), 10000);
    printf("%f\n", time2);
    double time3 = get_run_time(&_strlen3, a.c_str(), 10000);
    printf("%f\n", time3);
    double time4 = get_run_time(&strlen, a.c_str(), 10000);
    printf("%f\n", time4);
    double time5 = get_run_time(&_strlen4, a.c_str(), 10000);
    printf("%f\n", time5);
}