🚀 go-pugleaf

Bob Gonder <notbg@notmindspring.invalid> wrote:

>I disagree. The alignment should be inconsequential.
>If CPU A is doing a mov  (one cpu instruction) on 2 dwords (due to
>misalignment), then those 2 dwords should be locked by the memory
>handler (as it does an internal read/modify/write), so that CPU B
>can't access them.

Even if the two parts are in two different pages, and the second causes
a page fault? Or will the memory handler lock cause the page fault and
subsequent frantic activity as the page gets loaded?

>Only case where it should be a problem is if the cpu has to update the
>value in two parts, as in an int64.

Yes, for those. (And some may well consider an __int64 as a primitive.)

>That is the current state in single-cpu86-land.
>Interrups (other than instruction faults) occur between instructions.
>LOCK is used to lock memory (and interrupts) for what the book calls
>"indivisible read/modify/write cycles" (where the data physicaly
>leaves and reenters the memory module).

OK. I've not read the Intel docs on this.

>No, except for internal faults, the current instruction is completed
>before each interrupt. Other processors, however, may access the same
>memory during the course of another processor's instruction (as in
>INC). Thus the LOCK modifier.  Single-instruction stores such as MOV
>however, must be "safe" or else Intel wouldn't have made LOCK illegal
>on them (apparently, it was legal until the 386 came out).

Interesting. I suppose that was the sane course of action.

>>However, if I remember my assembler, there are some instructions that go
>>on for a potential lifetime by CPU clock cycle standards, so maybe it
>>does.
>
>That's one of the reasons we don't have "real" real-time os's.
>DIV is the longest I know of (other than FP, but they don't count)  at
>50 to 100 cycles. But BOUNDS, IRET with taskswitch, run 2-300, but I'd
>think the'yd be at the start or end of an interrupt, and not be
>subjected to further interrupts.

Isn't there a REP instruction that continues until CS or something
reaches zero?

Alan Bellingham
-- 
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