The patch that was posted on the review board was along the lines of the
solution suggested in my previous mail. Note that only splitting the
ccFlags register is not the complete solution. We would still need to
check that whether or not all the flags in a particular register are being
completely written or not. Otherwise two 'add' instructions would still
not get done in parallel.
But Gabe is not happy with the current state of the patch, so I dug in to
the isa_parser and here is another proposal. In operands.isa, let's mark
the ccFlagBits so that we can identify it in isa_parser. When build the
operand map, we can now figure out that we have an operand that may be
read conditionally (similar to how the PC is read and updated). In
functions makeRead(), makeConstructor functions, if we encounter such an
operand, we include a call to a check function which decides whether or
not this operand will be read. The check function will need the ext (or
the flag bits) to figure out if the operand needs to be read.
How does this sound?
--
Nilay
On Tue, 10 Apr 2012, Nilay Vaish wrote:
Gabe, Thanks for the explanation.
I think I have a solution for the problem that will work when all the flag
bits will be written. I am assumping that the condition code register is
identified as a source because it appears on the right hand side of the
following code in src/arch/x86/isa/microops/regop.isa
class FlagRegOp(RegOp):
abstract = True
flag_code = \
"ccFlagBits = genFlags(ccFlagBits, ext, result, psrc1, op2);"
Suppose we want to remove the RAW dependence in case of 'add' microop. We can
create a new microop with a different mnemonic. This new microop will be used
in places where currently 'add' is in use with all the flags. This microop
will not inherit from FlagRegOp class as 'add' does. It will inherit from a
class that assumes a default value for ccFlagBits. We will also need to add a
default value of 0 for ccFlagBits in the prototype of function genFlags().
But this will only help in case of addition and subtraction, as those are the
only ones that write all the six flag bits. As Gabe pointed out, for any good
solution to work, we would need to break up the flag bits.
--
Nilay
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