3 ms·
If you assume each process needs, say, 16MB of contiguous space, then you get 48-24 bits left, which by the birthday paradox implies you can have up to 2^(24/2
by smallnamespace 7y ago
If you assume each process needs, say, 16MB of contiguous space, then you get 48-24 bits left, which by the birthday paradox implies you can have up to 2^(24/2 = 12) ~= 4k processes before you start colliding about half the time.
- saagarjha 7y ago> If you assume each process needs, say, 16MB of contiguous space Unfortunately I’m not sure that’s a good assumption, due to the stack and heap needing to exist even for statically-linked binaries.
- smallnamespace 7y agoYes, but simply replace with 'average number of allocations * number of processes' and * 'average size of allocation'