initialize random udp package id using nanosecond timestamp #1270
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| unsigned long xorshf96(unsigned long& x, unsigned long& y, unsigned long& z) // period 2^96-1 | ||
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| const std::lock_guard<std::mutex> lock(xorshf96_mtx); | ||
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Why do you have a mutex here? what do you want to protect? This function has no global variables which it might access, hence it is reentrant.
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Yes that is right :)
KerstinKeller
approved these changes
Nov 24, 2023
| static std::mutex xorshf96_mtx; | ||
| const std::lock_guard<std::mutex> lock(xorshf96_mtx); | ||
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| static unsigned long x = static_cast<unsigned long>(std::chrono::duration_cast<std::chrono::nanoseconds>( |
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@rex-schilasky I was wondering if std::random would be a bit more appropriate here, to indicate we need a random number to start with. If you started two processes at the exact same time, would they get the same numbers this way? Though still unlikely.
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Description
Random UDP package id now uses timestamp to start with a unique number for every process.
Cherry-pick to