Lab 15.5: Implementing recursive functions in our VM

The goal of this lab is to practice implementing a virtual machine to interpret a pseudo-assembly language.

1. Introduction

The new model of our virtual machine is out--- its main new feature is language-level support for fuction calls. Here are the concepts behind it:

• The call stack is stored in main memory.
• The return address, frame pointer, and stack pointer are all stored in designated registers.
• A "local variable" is considered to be an offset within the stack frame.

When a function is called, the caller must

1. Store registers it wants to save into local variables
2. Store the old return address register into a local variable
3. "Push" a new frame on the stack (ie, adjust the frame pointer and stack pointer)
4. Store the old frame pointer in a local variable
5. Store the parameter in a local variable in the new frame.

In the new model, there exactly 32 registers. The last three have a special purpose:

• r29 The return address
• r30 The frame pointer
• r31 The stack pointer

These can be manipulated like the other registers, but some of the new instructions read or modify them without them being stated. Here are the new instructions, with their numeric codes:

• JAL r1 (12) Jump to instruction referred to by r1, storing the index of the next instruction (the one following this current JAL instrction) in the return address pointer.
• RET (13) Jump to the instruction referred to be the return address pointer.
• PSH n r1 (14) Push a new stack frame of size n onto the call stack; specifically, increment the stack pointer by n, set the frame pointer to be the old stack pointer, and save the old frame pointer in r1.
• POP r1 (15) Pop the top stack frame. Specifically, set the stack pointer to the old frame pointer, and restore the frame pointer to the value stored in r1.
• RELO n r1 (16) Load the local variable in memory position n (that is, n positions off from the beginning of the stack frame) into r1.
• WRLO n r1 (17) Write the value in r1 to the local variable in memory position n.

There is one other change. Now the second value in the file does not refer to the number of registers (since that's fixed in the new model) but rather it gives the initial value of the stack pointer-- where the stack frame for the "main" function ends. The frame pointer begins at zero---although the program could modify it to be something else, if there is a desire to differentiated between global addresses and the beginning of the call stack.

2. Setup

Make a new directory for this lab and copy the file from the appropriate course directory.

cp /cslab/class/cs245/labfifteenpt5/* .

You will find the source code for the version of the virtual machine from class on friday, a (compiled) disassembler for the new version, and a program written for the new version (unknown).

3. The unknown program

Disassemble the program unknown. Can you tell what it does? Take a couple mintues to try to decipher it (you may want to make a hard copy); if you don't make an progress, then move on. Once you run the program, it might be easier to make sense of it.

4. The new virtual machine

Your main activity will be to update the virtual machine program to run programs in the expanded language. You will need to

• Change the way you make the registers array.
• Initialize the designated registers (29-31). The return address can be undefined initially. (How do you initialize the stack pointer?)
• Implement the six new instructions.

Then test your program on the given program. If you still don't know what it does, try changing the first load to be something other than 5-- say, 4 or 6 or 7. (This will mean changing the fourth line of the file unknown.)

5. If you have time...

Try writing an assembly program of your own. A fairly simply one would be something to compute the greatest common divisor. Remember gcd(a, 0) = a, and gcd(a, b) = gcd(b, a % b) if b is not 0.

You'll probably want to write the program in phases. When I wrote unknown, I started by using "labels" instead of instruction positions for jumps.

        LOAD 5 r1
        LOAD 0 r2
        LOAD 1 r3
        LOAD L1 r4
        LOAD L2 r6
        PSH 4 r5
        WRLO 0 r1
        JAL r4
        POP r2
        RELO 0 r5
        PRN r5
        HLT
L1:
        RELO 0 r5
        IF r5 r6
        WRLO 0 r3
        RET
L2:
        SUB r5 r3 r7
        WRLO 1 r5
        WRLO 2 r29
        PSH 4 r8
        WRLO -1 r8
        WRLO 0 r7
        JAL r4
        RELO 0 r7
        RELO -1 r8
        POP r8
        RELO 2 r29
        RELO 1 r5
        MUL r5 r7 r8
        WRLO 0 r8
        RET

Then I counted off positions to determine what the labels where, and also replaced all the instructions and registers with plain numbers. Furthermore, I counted the number of instruction words.

84
        1 5 1
        1 0 2
        1 1 3
        1 31 4
        1 41 6
        14 4 5
        17 0 1
        12 4
        15 2
        16 0 5
        8 5
        9
L1:
        16 0 5
        7 5 6
        17 0 3
        13
L2:
        4 5 3 7
        17 1 5
        17 2 29
        14 4 8
        17 -1 8
        17 0 7
        12 4
        16 0 7
        16 -1 8
        15 8
        16 2 29
        16 1 5
        5 5 7 8
        17 0 8
        13

Finally, I made every number appear on a line by itself (and I got rid of the labels).

84
0
1 
5 
1
1 
0 
2
1 
1 
3
1 
31 
4
1 
41 
6
14 
4 
5
17 
0 
1
12 
4
15 
2
16 
0 
5
8 
5
9
16 
0 
5
7 
5 
6
17 
0 
3
13
4 
5 
3 
7
17 
1 
5
17 
2 
29
14 
4 
8
17 
-1 
8
17 
0 
7
12 
4
16 
0 
7
16 
-1 
8
15 
8
16 
2 
29
16 
1 
5
5 
5 
7 
8
17 
0 
8
13

6. To turn in

Cat all your files in a typescript which also shows you running the programs. Turn in a hard copy.