In Banker's algorithm of deadlock avoidance, if one can obtain a safe state, does that necessarily mean there will never be a deadlock, or is there still some way that the system could fall into a deadlock even if a safe state exists?

  • $\begingroup$ Safe state means there exists some resource allocation scheme which does not lead to a deadlock i.e. if you choose such an allocation, you won't get a deadlock state. Performing any other allocation might lead to a deadlock, but not necessarily. $\endgroup$ – RandomPerfectHashFunction Oct 3 '19 at 20:10

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