5 ms·
In the case of a traditional filesystem, similarly unfortunate memory corruption would just affect the actual data directly. Forgoing ZFS throws out the happy p
by mclehman 7y ago
In the case of a traditional filesystem, similarly unfortunate memory corruption would just affect the actual data directly. Forgoing ZFS throws out the happy path where the checksum does catch a data error and repairs it transparently.
Edit: To add on to this, ZFS doesn't do parity-based repair at the checksum level. In the mentioned case (good data, bad checksum) no copy of the block will match the checksum and there will be no supposedly-good block to copy over.