This PR adds a test case containing a Wishbone subordinate. The Makefile, C++ testbench and Verilog files (in the interpreter's test for this example) are taken from:

• https://zipcpu.com/tutorial/lsn-04-pipeline.pdf
• https://zipcpu.com/zipcpu/2017/05/29/simple-wishbone.html

(with minor modifications so that Verilator works on M-series Macs)

The VCD file is produced by their test-bench.

The monitor infers the following trace from the VCD:

// trace 0
trace {
    read(0, 0);
    write(0, 0);
    read(0, 3);
    read(0, 5);
    read(0, 7);
    read(0, 9);
    read(0, 11);
    read(0, 0);
    write(0, 0);
    read(0, 3);
    read(0, 5);
    read(0, 7);
    read(0, 9);
    read(0, 11);
    read(0, 0);
}

This mostly matches the .tx file supplied to the interpreter (the only difference is that the two writes in the .tx file are write(0, 1) instead of write(0, 0)). There are some differences between the interpreter and their test-bench which I've been unable to figure out, so we should exclude this example from the round-trip tests.

Side-note about stall and ack

Note: in the write protocol, we only wait till ack becomes 1, like so:

// Wait until Ack becomes 1 
// (`stall` becomes 1 in the same cycle as `ack`, so we don't wait for it in a separate loop) 
while (DUT.o_ack == 1'b0) {
    step();
} 
...

We don't need to wait for stall to become 0 in a separate while-loop, because the spec for Wishbone B4 Pipeline mode allows for ack and stall to change at the same time. In this particular Verilog DUT, stall and ack are both 1 in the same cycle (the ack = 1 means the current request has been serviced, whereas stall = 1 indicates that the next request has to wait).

If instead we had written the following, this would be wrong:

// Wait for stall to become 0, then wait for ack to become 1
// This is wrong, since `stall` and `ack` can be updated at the same time
while (DUT.o_stall == 1'b1) { 
  step(); 
}
while (DUT.o_ack == 1'b0) { 
  step(); 
}

The reason why this is wrong is that since stall and ack can be updated at the same time, if we waited for them sequentially (one after another), we would miss a ack = 1 that could have arrived when we still had stall = 1.

Thus, in the write protocol in this PR, we only wait for ack to become 1.