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Excerpts from mi400: 23-Nov-99 stack vs. storage-to-storage Gene Gaunt@ibm.net (433) > Why is the stack virtual machine better than the storage-to-storage > virtual machine? Is it faster? The key advantage to a stack-based virtual machine is in code optimization. Consider this simple sequence of HLL code: x = (a * a) + (b * b); y = x + (c * c); In the storage-to-storage model of Original MI the code expansion might look like: MULT T1, A, A <--updates storage MULT T2, B, B <--updates storage ADDN X, T1, T2 <--updates storage MULT T1, C, C <--updates storage ADDN Y, X, T3 <--updates storage The 'storage-to-storage' Original MI architecture requires that T1 and T2 -- the two required temporary variables -- be updated by the MULT instructions. This means storing the intermediate results "all the way home" to their locations in the automatic or static storage frame. Now consider the stack-based machine expansion, in pseudo-code: LOD A DUP MULT LOD B DUP MULT ADD STR X <-- updates storage LOD X LOD C DUP MULT ADD STR Y <-- updates storage While there are more 'pseudo' instructions, note that the only places that storage is updated is to actually update the result locations. All processors wait at the same speed... One of the things that processors have to wait for is the time it takes to get items to and from "storage." The "distance" to storage, compared to the speed of the processor, has been a critical source of performance problems for years. Partial solutions such as 'data caching' -- effectively keeping some of the storage logically 'closer' to the processor -- have been very successful, but do not solve the fundamental problem. Further, they suffer from problems of their own, such as keeping the cache consistent when there are multiple processors. Therefore, avoiding trips to "storage" -- especially distant storage -- is a key facilitator for optimization. Now, the 'stack' in the virtual machine can actually map to processor registers -- the absolute "closest" of the close storage -- and so we can get this simple sequence to happen a WHOLE lot faster. Furthermore, if it can be determined that 'x' is never used again in the program, it need never even *have* a final storage location, instead living its entire life in a single register. Anyway, that's probably more than you wanted, but it's a peek at just one of the architectural advantages of a stack-based machine -- the 'locality' of the logical stack storage and the avoidance of long trips to storage. HTH. -blair +--- | This is the MI Programmers Mailing List! | To submit a new message, send your mail to MI400@midrange.com. | To subscribe to this list send email to MI400-SUB@midrange.com. | To unsubscribe from this list send email to MI400-UNSUB@midrange.com. | Questions should be directed to the list owner/operator: dr2@cssas400.com +---
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