What does `~0>>-3` mean?

Question

#define BITMAP_LAST_WORD_MASK(nbits) (~0UL >> (-(nbits) & (BITS_PER_LONG - 1)))

This is the operation of bitmap in Linux. nbits is 3 and BITS_PER_LONG is 64. I thought the expression should be ~0UL >> -3 and the result means I have three items. In the expression I do the & first and add -. Maybe the expression has been wrong already.

Answer 1

Assuming nbits is nonnegative, -(nbits) & (BITS_PER_LONG - 1) computes BITS_PER_LONG - nbits%BITS_PER_LONG except that, if nbits is a multiple of BITS_PER_LONG, it produces zero instead of BITS_PER_LONG.

~ inverts all the bits in its operand, so ~0UL is a long with all its bits set. Shifting this right by BITS_PER_LONG - nbits%BITS_PER_LONG shifts zeros into the high bits, leaving only the low nbits set to one. Since the expression shifts by the & expression rather than BITS_PER_LONG - nbits%BITS_PER_LONG, if nbits is zero, it does not shift at all, leaving all bits set.

Thus the result of (~0UL >> (-(nbits) & (BITS_PER_LONG - 1))) is:

• if nbits is a multiple of BITS_PER_LONG, an unsigned long with all bits set,
• otherwise, an unsigned long with the low nbits%BITS_PER_LONG bits set.