Packets coming over a network have padding bytes added at the end for alignment. I want to skip these bytes but the packet size is variable but known. Given a number n, how do I round it up to the next 4-byte alignment?
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template boy
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http://stackoverflow.com/questions/4840410/how-to-align-a-pointer-in-c – simon Apr 28 '15 at 17:08
2 Answers
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For any integer n and any stride k (both positive), you can compute the smallest multiple of k that's not smaller than n via:
(n + k - 1) / k * k
This uses the fact that integral division truncates.
Kerrek SB
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@templateboy: If you're counting in bits, yes, though that's pretty unlikely. You should probably count in bytes, i.e. in the size of the smallest allocatable (and hence alignable) unit of storage. – Kerrek SB Apr 28 '15 at 17:08
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Another version. n is the number you want to alight to 4 (say k). Formula would be=n+k-n%k (where % is modulus)
For example (in Unix bc calculator)
k=4
n=551
n+k-n%k
552
to check that it is aligned:
scale=4
552/4
138.0000
Madars Vi
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4if n is already aligned this adds an extra k: e.g. n=4; k=4; 4+4-(4%4)= 8, not 4. it works, if you check n%k != 0 first, but in some code there are penalties for taking a branch and @Kerrek SB's solution is preferred. – Brad May 31 '20 at 00:31