To: pgerassi who wrote (74101 ) 3/9/2002 3:16:39 AM From: Petz Read Replies (2) | Respond to Register renaming does not store temporary results in a program. Those results must be written to L1 cache, then L2 cache and finally to memory. The CPU can not make assumptions about future code not read yet. could regognize) that it need not load from the memory location because the value is still in one of the copies of the EAX register. Maybe I am wrong about that -- how would the CPU know whether the temp variable is in a DMA buffer which could invalidate the copy in the register. (A compiler would know because the variable would be declared 'volatile'.) -- unless it goes through the mechanics of doing a memory read, in which case any DMA should have invalidated the cache copies of the data and forced another memory read. And you are right that a compiler designed for 64-bit mode would be aware of more "original registers" would usually allow it to substitute a register for the TEMPxx memory variables in my example. Register renaming is used only when a register will hold a result which may or may not be overwritten. Thus there may be a copy of EAX for each of the stages within the execution pipeline. If thats 6 stages then you may have 7 copies (the original plus a copy at the end of each stage) in 7 different virtual registers. The [function] calls are helped because more registers are available to hold parameters instead of them being on the stack with all of the attended cache costs. All in all, 64 bit "long" mode will cause many compilers to generate better faster more optimal code.
