Discussion:
HW6: 5.2/5.3 - ALUOp
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Jeremy Cowles
2008-08-01 17:43:02 UTC
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When it says ALUOp0 and ALUOp1, I made the assumption that this meant the
the 0 or 1 bit in the 2 bit ALUOp signal. So when the signal is stuck at
zero:

ALUOp0 >> ALUOp = x0
ALUOp1 >> ALUOp = 0x

And when it is stuck at 1:

ALUOp0 >> ALUOp = x1
ALUOp1 >> ALUOp = 1x

Where "x" is either 1 or 0 depending on the instruction. Is this correct?

Thanks,
Jeremy
Jeremy Cowles
2008-08-01 19:38:23 UTC
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Never mind, this was answered under Neil's post.
Post by Jeremy Cowles
When it says ALUOp0 and ALUOp1, I made the assumption that this meant the
the 0 or 1 bit in the 2 bit ALUOp signal. So when the signal is stuck at
ALUOp0 >> ALUOp = x0
ALUOp1 >> ALUOp = 0x
ALUOp0 >> ALUOp = x1
ALUOp1 >> ALUOp = 1x
Where "x" is either 1 or 0 depending on the instruction. Is this correct?
Thanks,
Jeremy
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