Hi Christian, It took me a while to get my head around this. I'm not convinced that the documentation is incorrect, although they are maybe looking at it from a slightly different perspective. Keep in mind Figure 6-3... the current partition is above the dirty partition is above the clean partition. RSE operation should preserve this relationship (as I see it). Also, note that the registers in the current partition, even though not technically part of the dirty partition, are necessarily dirty... they may be targeted at any time by writebacks. > while developing an instruction set simulator I'd be interested in learning more about this simulator :) > incomplete register frame (ASDM vol. 2) to be inaccurate. It says that > the incomplete frame becomes complete when RSE.ndirty_words becomes > non-negative by executing mandatory RSE loads. However, mandatory RSE > loads are not touching BSPSTORE or the dirty partition. Indirectly they are, because presumably the dirty partition (and BSPSTORE) must remain squeezed between the invalid partition and the current partition (albeit being zero-size, or negative-size depending on how you look at it - the invalid partition extends into the current partition!). > RSE loads are extending the clean partition. IMHO there is no clean partition. The RSE loads are targeting the current frame so they are immediately dirty+current. > Furthermore if you consider RSE.BOF to > end up inside the clean partition after a br.ret or rfi, then BSPSTORE > is located above BSP. RSE.BOF by definition ends up in the current partition (which is immediately dirty), which is above the other dirty registers (if any) in the dirty partition, which is above the clean partition. Hence in the normal case BSPSTORE is less than or equal to BSP. Hope this helps, feel free to argue with me if you disagree. Matt - : send the line "unsubscribe linux-ia64" in the body of a message to majordomo@xxxxxxxxxxxxxxx More majordomo info at http://vger.kernel.org/majordomo-info.html
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