Scheduling question.

R. Clayton (rclayton@monmouth.edu)
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  Since the initial and final values don't matter, b+c would be translated into
  the CPU instructions

    move r0 b
    move r1 c
    + b c

Ah, careful. I didn't say the values didn't matter; I said they were
arbitrary. The difference is the same as the difference between "the value x"
and "the value 5".

   Because the value of b + c isn't being stored, would it be correct if my
   scheduler comes up with no schedule at all?

I don't know. Are the results (in storage) of the three-instruction schedule
the same as the results (in storage) of a zero-instruction schedule? If so,
then it's correct; otherwise, it's not correct.

But this assumes that the third instruction is "+ r0 r1". As written, the
third instruction is syntatically incorrect (op arguments must be registers).

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