if (__builtin_expect (end_ptr < str, 0)) end_ptr = (const char *) ~0UL;
__strnlen (const char *str, size_t maxlen){  const char *char_ptr, *end_ptr = str + maxlen;  const unsigned long int *longword_ptr;  unsigned long int longword, magic_bits, himagic, lomagic;  if (maxlen == 0)    return 0;  /* 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')      {	if (char_ptr > end_ptr)	  char_ptr = end_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.  */  magic_bits = 0x7efefeffL;  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.  */      magic_bits = ((0x7efefefeL << 16) << 16) | 0xfefefeffL;      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.  */  while (longword_ptr < (unsigned long int *) end_ptr)    {      /* We tentatively exit the loop if adding MAGIC_BITS to	 LONGWORD fails to change any of the hole bits of LONGWORD.	 1) Is this safe?  Will it catch all the zero bytes?	 Suppose there is a byte with all zeros.  Any carry bits	 propagating from its left will fall into the hole at its	 least significant bit and stop.  Since there will be no	 carry from its most significant bit, the LSB of the	 byte to the left will be unchanged, and the zero will be	 detected.	 2) Is this worthwhile?  Will it ignore everything except	 zero bytes?  Suppose every byte of LONGWORD has a bit set	 somewhere.  There will be a carry into bit 8.  If bit 8	 is set, this will carry into bit 16.  If bit 8 is clear,	 one of bits 9-15 must be set, so there will be a carry	 into bit 16.  Similarly, there will be a carry into bit	 24.  If one of bits 24-30 is set, there will be a carry	 into bit 31, so all of the hole bits will be changed.	 The one misfire occurs when bits 24-30 are clear and bit	 31 is set; in this case, the hole at bit 31 is not	 changed.  If we had access to the processor carry flag,	 we could close this loophole by putting the fourth hole	 at bit 32!	 So it ignores everything except 128's, when they're aligned	 properly.  */      longword = *longword_ptr++;      if ((longword - lomagic) & himagic)	{	  /* 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);	  char_ptr = cp;	  if (cp[0] == 0)	    break;	  char_ptr = cp + 1;	  if (cp[1] == 0)	    break;	  char_ptr = cp + 2;	  if (cp[2] == 0)	    break;	  char_ptr = cp + 3;	  if (cp[3] == 0)	    break;	  if (sizeof (longword) > 4)	    {	      char_ptr = cp + 4;	      if (cp[4] == 0)		break;	      char_ptr = cp + 5;	      if (cp[5] == 0)		break;	      char_ptr = cp + 6;	      if (cp[6] == 0)		break;	      char_ptr = cp + 7;	      if (cp[7] == 0)		break;	    }	}      char_ptr = end_ptr;    }  if (char_ptr > end_ptr)    char_ptr = end_ptr;  return char_ptr - str;}