Memory alignment # DEFINE _ intsizeof (N) (sizeof (n) + sizeof (INT)-1 )&~ (Sizeof (INT)-1 ))

For two positive integers x and N, the total number of integers Q, R makes

X = NQ + R, where 0 <= r <N // minimum non-negative residue

Q and R are uniquely identified. Q = [x/n], r = x-N [x/n]. This is a simple form with remainder division. In C, Q and R are easy to calculate: q = x/N, r = x % N.


Alignment of X by N refers to: If r = 0, take Qn. If R> 0, take (q + 1) n. This is equivalent to representing X as follows:

X = NQ + R', where-n <R' <= 0 // The maximum is not positive.

NQ is what we want. The key is how to calculate it in C language. Since we can process the standard Division with remainder, we can convert this formula into a standard Division with remainder, and then process it:

X + n = qn + (n + R'), where 0 <n + R' <= N // The largest positive Remainder

X + n-1 = qn + (n + R'-1), where 0 <= N + R'-1 <N // minimum non-negative residue

Therefore, qN = [(x + N-1)/n] n. Calculated in C language:

(X + N-1)/n) * n

If n is the power of 2 For example, 2 ^ m is used to remove the right M bit, and the multiplication is used to shift the left M bit. So we can clear the minimum M binary bits of X + n-1 to 0. Get:

(X + N-1 )&(~ (N-1 ))