How does this macro detect alignment issues?

Question

I am tracing some source code about the implementation of strlen:

#include <_ansi.h>
#include <string.h>
#include <limits.h>

#define LBLOCKSIZE   (sizeof (long))
#define UNALIGNED(X) ((long)X & (LBLOCKSIZE - 1))

#if LONG_MAX == 2147483647L
#define DETECTNULL(X) (((X) - 0x01010101) & ~(X) & 0x80808080)
#else
#if LONG_MAX == 9223372036854775807L
/* Nonzero if X (a long int) contains a NULL byte. */
#define DETECTNULL(X) (((X) - 0x0101010101010101) & ~(X) & 0x8080808080808080)
#else
#error long int is not a 32bit or 64bit type.
#endif
#endif

#ifndef DETECTNULL
#error long int is not a 32bit or 64bit byte
#endif

size_t
_DEFUN (strlen, (str),
    _CONST char *str)
{
_CONST char *start = str;

#if !defined(PREFER_SIZE_OVER_SPEED) && !defined(__OPTIMIZE_SIZE__)
unsigned long *aligned_addr;

/* Align the pointer, so we can search a word at a time.  */
while (UNALIGNED (str))
    {
    if (!*str)
    return str - start;
    str++;
    }

/* If the string is word-aligned, we can check for the presence of
    a null in each word-sized block.  */
aligned_addr = (unsigned long *)str;
while (!DETECTNULL (*aligned_addr))
    aligned_addr++;

/* Once a null is detected, we check each byte in that block for a
    precise position of the null.  */
str = (char *) aligned_addr;

#endif /* not PREFER_SIZE_OVER_SPEED */

while (*str)
    str++;
return str - start;
}

Most of the code I can understand, but I have no idea how the following macro can tell if a string is word-aligned:

#define UNALIGNED(X) ((long)X & (LBLOCKSIZE - 1))

How does it work?

Answer 1

If LBLOCKSIZE is a power of two, than LBLOCKSIZE - 1 is a pattern of all 1s in the low bits. If any bit in that pattern is set in an address, then it is not aligned to that blocksize.

Example:

LBLOCKSIZE = 4096
LBLOCKSIZE - 1 = 4095 = 0xFFF

Typically, disk blocks and memory blocks are power-of-two sizes because hardware tends to prefer them. Although some of us are old enough to remember decimal machines.