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We implement persistent pointers in a persistent programming language by using OIDs In some implementations, persistent pointers are physical OIDs; in others, they are logical OIDs An important difference between persistent pointers and in-memory pointers is the size of the pointer In-memory pointers need to be only big enough to address all virtual memory On most current computers, in-memory pointers are usually 4 bytes long, which is suf cient to address 4 gigabytes of memory The most recent computer architectures have pointers that are 8 bytes long, Persistent pointers need to address all the data in a database Since database systems are often bigger than 4 gigabytes, persistent pointers are usually at least 8 bytes long Many object-oriented databases also provide unique identi ers in persistent pointers, to catch dangling references This feature further increases the size of persistent pointers Thus, persistent pointers may be substantially longer than in-memory pointers The action of looking up an object, given its identi er, is called dereferencing Given an in-memory pointer (as in C++), looking up the object is merely a memory reference Given a persistent pointer, dereferencing an object has an extra step nding the actual location of the object in memory by looking up the persistent pointer in a table If the object is not already in memory, it has to be loaded from disk We can implement the table lookup fairly ef ciently by using a hash table data structure, but the lookup is still slow compared to a pointer dereference, even if the object is already in memory

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The equivalent resistance is (R1 + R3 ) (R2 + R4 ) 1 1 1 = + RTH = RTH R1 + R3 R2 + R4 R1 + R3 + R2 + R4 (2 + 6) (3 + 6) = = 42 2+6+3+6 Remember, since each combination shown in Fig 3-22 is in series, we add them together to get the equivalent resistance We can then replace the complex circuit with the resistors connected in a diamond shape (Fig 3-19) by the Thevenin equivalent circuit shown in Fig 3-23, which is much simpler Now we can nd the voltage in the load resistor Reconnecting it, we need to solve the circuit shown in Fig 3-24 Applying KVL to the loop, we have VTH + RTH I + VL = 0 VL = VTH RTH I = 125 (3)(42) = 114 V

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