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sander.saares.eu | ||
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danlark.org
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| | | | | When it comes to hashing, sometimes 64 bit is not enough, for example, because of birthday paradox -- the hacker can iterate through random $latex 2^{32}$ entities and it can be proven that with some constant probability they will find a collision, i.e. two different objects will have the same hash. $latex 2^{32}$ is around... | |
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www.bazhenov.me
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| | | | | Introduction Link to heading Suppose we need to write a function that computes the next set of numbers in a range and stores them in a slice, as shown below: let pl = RangePl::new(1..12); let mut buffer = [0u64; 4]; pl.next_batch(0, &mut buffer); // returns 4, buffer[..] = [1, 2, 3, 4] pl.next_batch(0, &mut buffer); // returns 4, buffer[..] = [5, 6, 7, 8] pl.next_batch(10, &mut buffer); // returns 2, buffer[0..2] = [10, 11] pl.next_batch(10, &mut buffer); // returns 0, buffer not updated Key factors: | |
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orlp.net
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blog.atx.name
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| | | atx - My personal blog | ||
