size_t
strlen(const char *str)
{
const char *p;
const unsigned long *lp;
/* Skip the first few bytes until we have an aligned p */
for (p = str; (uintptr_t)p & LONGPTR_MASK; p++)
if (*p == '\0')
return (p - str);
/* Scan the rest of the string using word sized operation */
for (lp = (const unsigned long *)p; ; lp++)
if ((*lp - mask01) & mask80) {
p = (const char *)(lp);
testbyte(0);
testbyte(1);
testbyte(2);
testbyte(3);
#if __WORDSIZE == 64
testbyte(4);
testbyte(5);
testbyte(6);
testbyte(7);
#endif
}
/* NOTREACHED */
return (0);
}
size_t
strlen(const char *str)
{
const char *p;
const unsigned long *lp;
long va, vb;
/*
* Before trying the hard (unaligned byte-by-byte access) way
* to figure out whether there is a nul character, try to see
* if there is a nul character is within this accessible word
* first.
*
* p and (p & ~LONGPTR_MASK) must be equally accessible since
* they always fall in the same memory page, as long as page
* boundaries is integral multiple of word size.
*/
lp = (const unsigned long *)((uintptr_t)str & ~LONGPTR_MASK);
va = (*lp - mask01);
vb = ((~*lp) & mask80);
lp++;
if (va & vb)
/* Check if we have \0 in the first part */
for (p = str; p < (const char *)lp; p++)
if (*p == '\0')
return (p - str);
/* Scan the rest of the string using word sized operation */
for (; ; lp++) {
va = (*lp - mask01);
vb = ((~*lp) & mask80);
if (va & vb) {
p = (const char *)(lp);
testbyte(0);
testbyte(1);
testbyte(2);
testbyte(3);
#if (LONG_BIT >= 64)
testbyte(4);
testbyte(5);
testbyte(6);
testbyte(7);
#endif
}
}
/* NOTREACHED */
return (0);
}
