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Fix maintenance time computation #313

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theoreticalbts 2015-09-16 11:07:36 -04:00
parent 1451f67636
commit 81cc8e4d81
1 changed files with 19 additions and 4 deletions
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23
libraries/chain/db_maint.cpp
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@ -521,10 +521,25 @@ void database::perform_chain_maintenance(const signed_block& next_block, const g
next_maintenance_time = time_point_sec() + next_maintenance_time = time_point_sec() +
(((next_block.timestamp.sec_since_epoch() / maintenance_interval) + 1) * maintenance_interval); (((next_block.timestamp.sec_since_epoch() / maintenance_interval) + 1) * maintenance_interval);
else else
// It's possible we have missed blocks for at least a maintenance interval. {
// In this case, we'll need to bump the next maintenance time more than once. // We want to find the smallest k such that next_maintenance_time + k * maintenance_interval > head_block_time()
do next_maintenance_time += maintenance_interval; // This implies k > ( head_block_time() - next_maintenance_time ) / maintenance_interval
while( next_maintenance_time < head_block_time() ); //
// Let y be the right-hand side of this inequality, i.e.
// y = ( head_block_time() - next_maintenance_time ) / maintenance_interval
//
// and let the fractional part f be y-floor(y). Clearly 0 <= f < 1.
// We can rewrite f = y-floor(y) as floor(y) = y-f.
//
// Clearly k = floor(y)+1 has k > y as desired. Now we must
// show that this is the least such k, i.e. k-1 <= y.
//
// But k-1 = floor(y)+1-1 = floor(y) = y-f <= y.
// So this k suffices.
//
auto y = (head_block_time() - next_maintenance_time).to_seconds() / maintenance_interval;
next_maintenance_time += (y+1) * maintenance_interval;
}
} }
modify(get_dynamic_global_properties(), [next_maintenance_time](dynamic_global_property_object& d) { modify(get_dynamic_global_properties(), [next_maintenance_time](dynamic_global_property_object& d) {