If the PC register is simultaneously read and written, does its read data contain the previous data or the newly-written data?

Question

If the PC register is simultaneously read and written, does its read data contain the previous data or the newly-written data? Based on my understanding of sequential circuits, the effect of the write command does not instantly take effect in the PC register due to propagation delay so, at the rising edge of the clock, the read command will get the old value. But corollary to my question is if this is the case, shouldn't the read command would also have a delay in some sense and could possibly read the newly-written data?

Answer 1

A program counter is normally special enough that it's not part of a register file with other registers. You don't have a "read command", its output is just always wired up to other parts that read it when appropriate. (i.e. when its output is stable and has the value you want). e.g. see various block diagrams of MIPS pipelines, or non-pipelined single-cycle or multi-cycle designs.

You'd normally build such a physical register out of edge-triggered flip-flops, I think. (https://en.wikipedia.org/wiki/Flip-flop_(electronics)). Note that a D flip-flop does latch the previous input as the current output on a clock edge, and then the input is allowed to change after that.

There's a timing window before the clock edge where the input has to remain stable, it can start to change a couple gate delays after. Note the example of a shift register built by chaining D flip-flops all with the same clock signal.

If you have a problem arranging to capture a value before it starts changing, you could design in some intentional clock skew so the flip-flop reliably latches its input before you trigger the thing providing the input to change it. (But normally whatever you're triggering will itself have at least a couple gate delays before its output actually changes, hence the shift-register made of chained D flip-flops.)

That wiki article also mentions the master-slave edge-triggered D Flip-Flop that chains 2 gated (not clocked) D latches with an inverted clock, so capturing the input happens on the opposite clock edge from updating the output with the previously-captured data.

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By comparison and for example, in register files for general-purpose registers in classic RISC pipelines like MIPS, IIRC it's common to build them so write happens in the first half-cycle and read happens in the second half-cycle of the ID stage. (So write-back can "forward" to decode/fetch through the register file, keeping the window of bypass-forwarding or hazards shorter than if you did it in the other order.)

This means the write data has a chance to stabilize before you need to read it.

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Overall, it depends how you design it!

If you want the same clock edge to update a register with inputs while also latching the old value to the output, you a master-slave flip-flop will do that (capture the old input into internal state, and latch the old internal state onto the outputs).

Or you could design it so the input is captured on the clock edge, and propagates to the output after a few gate delays and stays latched there for the rest of this clock cycle (or half cycle). That would be a single D flip-flop (per bit).