GNU牛人写的strlen()

我不是女神ヾ 2022-08-24 09:04 206阅读 0赞

\#include <string.h>  
\#include <stdlib.h>  
  
\#undef strlen  
  
 /\*  
Return the length of the null-terminated string STR.  Scan for  
the null terminator quickly by testing four bytes at a time.    
\*/  
size\_t strlen (str)  
const char \*str;  
\{   
    const char \*char\_ptr;  
    const unsigned long int \*longword\_ptr;  
    unsigned long int longword, magic\_bits, 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.    
    \*/  
    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.    
    \*/  
    for (;;)  
    \{   
         /\* 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 (  
        \#if 0  
        /\* Add MAGIC\_BITS to LONGWORD.  \*/  
        (((longword \+ magic\_bits)  
  
        /\* Set those bits that were unchanged by the addition.  \*/  
        ^ ~longword)  
  
         /\* Look at only the hole bits.  If any of the hole bits  
        are unchanged, most likely one of the bytes was a  
        zero.  \*/  
        & ~magic\_bits)  
        \#else  
        ((longword \- lomagic) & himagic)  
        \#endif  
        != 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;  
            \}  
        \}  
    \}  
\}  
libc\_hidden\_builtin\_def (strlen)