Pages created and updated by Terry Sturtevant Date Posted: May 12, 2017

CP316: Microprocessor Systems and Interfacing

Introduction to the Development Environment (Debugger)

Objectives

When learning a new microprocessor or microcontroller, especially one with a Harvard architecture like the PIC, it is important to understand how the memory address space is structured. For microprocessors with a Harvard architecture, the data and program memory may have completely different structures. It is also important to determine how the IDE can be used to view memory and how the IDE will respond when you access nonexistent memory.
  1. understanding program memory addresses
  2. understanding data memory addresses
  3. introduce additional functions of the MPLABX IDE
  4. introduce use of the MPLABX IDE debugger
  5. understanding the operation of the banked memory

Equipment

Procedure

NOTE: Some groups may find this lab long. Any material not completed in the lab is homework.

The MPLABX debugger allows you to run code on the target board, while still being able to single step, use breakpoints, and examine memory.

  1. Watch this video on viewing memory in MPLABX. (It's about 6 minutes long.)

    First, you need to access project properties:

    project properties menu item

    To choose the debugger or programmer, you need to choose the ICD3 instead of the simulator in the project properties:
    project properties ICD3

    (Note: If the debugger is already running, it must be stopped in order to allow access to project properties.)

    To simplify using watch windows, you need to set to build in absolute mode.
    project properties absolute addresses
    After this, you should be able to use variable names when setting watches.

    Instead of using the menu, you can access project propertties using the wrench icon on the dashboard.
    icon for project properties
  2. The PIC18F452 has 32 Kbytes of Flash program memory, where K is 1024. How many bytes (in decimal) are contained in the Flash program memory?

  3. Program memory is byte addressable which means every byte has an address. If the lowest address is H'0', what is the highest address (in hex) in the Flash memory? Show your calculations.

  4. How many bits are required to represent this address range, i.e. how many bits are required to store the highest address? How does this number compare to the width of the instruction address bus as presented in Figure 1.3 on page 13 of the text?

  5. The PIC18F452 has the potential for addressing 2 Mbytes of program memory space with its 21-bit program counter. Not all models of the PIC18 family have the full 2Mbytes implemented on the microcontroller. Note that unimplemented program memory is always at the end of the program memory address space. If you access a nonexistent memory location between the physically implemented memory and the 2 Mbytes maximum, the processor will cause a read of all ’0’s (a NOP instruction).

    How much program memory does the MPLABX IDE let you view? Open the IDE, and look at program memory in hex. (Make sure you have the correct device selected under configuration. A program doesn't have to be loaded.) Does this match the implemented (32Kbytes) or the theoretical (2Mbytes) program memory range?

  6. The data memory is implemented as byte addressable static RAM. If each location in the data memory has a 12-bit address, and all bits are used, what is the lowest and highest address (in binary and in hex)? How many locations (in decimal and in Kbytes) are in data memory?

  7. The implemented data memory is less than the theoretical data memory. Unlike the program memory the unimplemented portion of the data memory space is in the middle of the address range. In the IDE, open a memory view of file registers and scroll through the address space to find the large gap of unimplemented memory. Note the valid address ranges disregarding the small gaps in high memory. Look at Figure 4-7 from the datasheet. Note what it says about "unused" memory.

  8. The lower chunk of data memory corresponds to General Purpose Registers (GPR) which is the general data area for the microcontroller. Show that this chunk corresponds to 1.5 Kbytes. In general, the GPR space starts at location H'000' and grows up in memory.

  9. The significantly smaller upper chunk of data memory corresponds to Special Function Registers (SFR) and these are used to control and get the status of the controller and peripheral functions. In general , the SFR space starts at location H'0FFF' and grows down in memory. In the IDE, if you view the file registers using the Symbolic option, you will notice the SFRs have predefined names. This area can also be viewed by selecting a view of special function registers in the IDE.

    Demonstration - explain the program and data memory layouts; explain how you can view these areas in the IDE.

  10. Set up a test project that contains your version of test.asm from the last lab. Assemble the program and open a window containing the listing file, test.lst. View program memory using symbolic mode. You should see the correspondence between the listing file and the contents of program memory. How does the display change if you select the or the hex option to view program memory?

  11. In the listing file, find the following lines of code:
    000018             org        0x0018        ;low priority interrupt vector
    000018 EF0C F000   goto       $             ; ... tight loop for now
    Find the corresponding location in the program memory. What was the  goto $  translated into?

    The "$" sign is a special character and it represents the address of the present instruction. In this section of code it is used to create a tight loop. If, by some chance, a low priority interrupt occurred, the program would go into a tight infinite loop. The program does not expect any interrupts and is using this section of code as a place holder.

  12. Switch to the file register view. Remember this is data memory and locations 0x0000 to 0x0002 should correspond to num1, num2, and result. Notice that under the symbolic option these locations are called A, AN1, and AN2. Since we didn't define these labels, they must be coming from the "include" file. Look at the symbol table at the end of the listing file. Did the include file labels overwrite our labels or do these memory locations have multiple labels? Does the file register view show the first label to be defined for a memory location or does it show the label that would be first if sorted alphabetically? Modify the program and test your hypothesis.

  13. What happens to the file register view if you comment out the include and the configuration lines? Note that if you only comment out the include, the configuration definitions will not work as they depend on labels defined in the include.

    Demonstration - explain the program memory view and the file register view.

  14. Go back to your version of test.asm from the last lab and rebuild. Launch the debugger, and ready the program for execution by selecting reset Note where the execution cursor position is located. Explain.

  15. Now single step through the program using the step into button. While stepping through the program notice the execution cursor position and the contents of the status line at the top of the screen.
    status bar
    • What does the status register contain at the start of the program?
    • What does the status register contain after completion of the add?
    • Does the current program counter contain the value of the line currently being executed or the next line to be executed? How can you tell?
    • Note the contents of the W register as you step through the program.

  16. Reset. Open a program memory view (symbolic) and a file register view (symbolic). Within the debugger, clear the file registers. (You can right click in a data memory view and Fill memory with zeros.) Now single step through the program.

  17. It is simple enough with only three variables to mentally map between our labeling and the system's labeling in the file register view. However, with more variables, it is more efficient to set up a watch window (see the Quick Start Guide). Set up a watch window with our three variables (from symbols) and wreg (from SFR). Clear the file registers. Single step through the program. Did you notice the address of wreg? What does this mean?

    Demonstration - demonstrate the use of a watch window and explain the location of wreg.

  18. Data memory doesn't quite work like typical linear memory. Change the cblock address to 0x080. Using the watch window as your debugging tool, single step the program. Other than the movement of the data to a different location in memory, the program should work as you would expect.

  19. Change the cblock address to 0x100. Using the watch window as your debugging tool, single step the program. The program does not appear to work.

  20. Open the file register view and the program memory view as well as the watch window. Clear the file registers. Single step the program. Does the program work? Where does the data appear to be located?

  21. Memory is divided into banks containing locations 0x000 to 0x0ff. The memory bank select register, BSR, points to the bank number. By default, BSR is zero. Add the line
            movlb    1      ;set memory bank
    to the start of the main routine. Add BSR to the watch window. Clear the file registers and single step the program. Things should now look OK. The movlb instruction moves the literal, in this case "1", to the bank select register.

  22. Test the following combinations, making sure to clear the file registers before executing, and use the file register view to determine the location of the data memory being updated. Open the program memory view. You do not have to single step the program; simply run the program and stop it when it goes into the last tight loop. What can you deduce from this exercise?

    cblock bsr Location of num1, num2, result according to watch window Memory locations actually being updated
    0x0000 0    
    0x0000 1    
    0x0000 2    
    0x0100 0    
    0x0100 1    
    0x0100 2    
    0x0200 0    
    0x0200 1    
    0x0200 2    

    Demonstration - explain the rules governing the bank selection register.
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