CPE 315
Winter 2008
Lab #4 FAQ

1. How do I see inside RAM?
   You can use the same technique as for the register file but it's not necessary.
   The test program will load all results from memory to a register; you can verify results in the register file.

2. Which instructions does the control unit in Appendix C support?
   This control unit is only partially complete. Its purpose is to show you how to build a control unit from PLAs.

3. Do we have to support negative offsets in the LW/SW?
   Yes.

4. How big should Data memory be?
   Big enough to run the test program and no bigger.
   Note that the Ram address bus must be exactly == log ₂ Ram size

5. How about some hints on designing the instructions that aren't part of Fig. 5.24?
   
   1. ADDI
      Control decodes the ADDI instruction and sets the control bits appropriately.
      Be sure to choose a clever ALUOp value (not 11)
      
   2. LUI
      The design was done in lecture on Feb. 21.
      
   3. SLTU
      The design was done in Lab Exercise #5
      

6. What is the geometry of the Control ROM?
   Input: 6 bits (less if you minimize). Do not send additional input signals to Control!
   Output: 16 bits, some unused
   Size: 64 words (less if you minimize).
   

7. What is the geometry of the ALU Control Rom?
   Input: 8 bits or less (you should minimize). Do not send additional signals to ALU Control! (such as Op Code)
   Output: 5 bits, including SLTU (declare an 8 bit bus and leave 3 unused)
   Size: 256 words if you're crazy enough not to minimize

8. What's the difference between ADD and ADDU?
   For this lab, there is no difference.
   In a MIPS cpu, an overflow during an ADD will generate an exception.
   

9. Can we use the unused code in ALUOp (11)?
   No. It complicates the design, isn't necessary and is explicitly disallowed.

10. What SIM component should be used for Data Memory?
    RAM. There is an example usage in the SIM apis.
    Watch out for the preconditions!

11. The J loop instruction in the test program appears incorrect.
    It is correct. Recall the two zeroes that must be appended to the address.
    So the J jumps to address 0x10 == loop.