C Language Study - 三大字元處理函數實現

標籤：strlen   strcpy   strcmp   

三大字元處理函數實現||||

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strlen , strcpy, strcmp.

Glibc:

首先出場的是strcmp：

/* Compare S1 and S2, returning less than, equal to or   greater than zero if S1 is lexicographically less than,   equal to or greater than S2.  */intstrcmp (const char *p1, const char *p2){  const unsigned char *s1 = (const unsigned char *) p1;  const unsigned char *s2 = (const unsigned char *) p2;  unsigned char c1, c2;  do    {      c1 = (unsigned char) *s1++;      c2 = (unsigned char) *s2++;      if (c1 == '\0')return c1 - c2;    }  while (c1 == c2);  return c1 - c2;}

strcpy:

/* Copy SRC to DEST.  */char *strcpy (char *dest, const char *src){  return memcpy (dest, src, strlen (src) + 1);}

strlen:

/* Return the length of the null-terminated string STR.  Scan for   the null terminator quickly by testing four bytes at a time.  */size_tstrlen (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;    }}    }}

xnuslibc:

C Language Study - 三大字元處理函數實現