In this assignment, you will implement a virtual machine (VM) known as the P-machine (PM/0). The P-machine is a stack machine with two memory stores: the stack which is organized as a stack in the upper end and a data area for global variables in the lower end. The stack segment contains data to be used by the PM/0 CPU. The other segment is the text segment, which is organized as an array and contains the instructions to be executed by PM/0 CPU. The PM/0 CPU has four registers. These registers are named base pointer (BP), stack pointer (SP), program counter (PC) and instruction register (IR). They will be explained in detail later on in this document.
The Instruction Set Architecture (ISA) of the PM/0 has 23 instructions and the instruction format is as follows: OP L M
Each instruction contains three components (OP, L, M) that are separated by one space.OP is the operation code.
L indicates the lexicographical level.
M depending of the operators it indicates:
A number (instructions: LIT, INC).
A program address (instructions: JMP, JPC, CAL).
A data address (instructions: LOD, STO)
The identity of the operator OPR
(e.g. OPR 0, 2 (ADD) or OPR 0, 4 (MUL)).
The list of instructions for the ISA can be found in Appendix A and B.
P-Machine Cycles
The PM/0 instruction cycle is carried out in two steps. This means that it executes two steps for each instruction. The first step is the Fetch, where the actual instruction is fetched from the text segment in memory. The second step is the Execute, where the instruction that was fetched is executed using the data-stack segment in memory. This does not mean the instruction is stored in the stack.
Fetch Step:
In the Fetch Step, an instruction is fetched from the text segment and is placed in the IR register (IR code[PC]). Then, the program counter is incremented by 1 to point to the next instruction to be executed (PC PC + 1).
Execute Step:
In the Execute Step, the instruction that was fetched is executed by the VM. The OP component that is stored in the IR register (IR.OP) indicates the operation to be executed. For example, if IR.OP is the ISA instruction OPR (IR.OP = 02), then the M component of the instruction in the IR register (IR.M) is used to identify the operator and execute the appropriate arithmetic or relational instruction.
PM/0 Initial/Default Values:
Initial values for PM/0 CPU registers:
SP = MAX_DATA-STACK_HEIGHT;
BP = SP 1;
PC = 0;
IR = 0;
Initial stack store values:
stack[0] =0; stack[1] =0; ; stack[n-1] =0;
Constant Values:
MAX_DATA-STACK_HEIGHT is 1000
MAX_CODE_LENGTH is 500
Assignment Instructions and Guidelines:
- The VM must be written in C and must run on Eustis. If it runs in your PC but not on Eustis, For us it does not run.
- Submit to Webcourses:
- A readme document indicating how to compile and run the VM.
- The source code of your PM/0 VM.
- The output of a test program running in the virtual machine. Please provide a copy of the initial state of the stack and the state of stack after the execution of each instruction. Please see the example in Appendix C.
- Student names must be shown at the beginning of the program.
- Include comments in your program.
- Do not implement each VM instruction with a function. If you do, a penalty of 50% will be applied to your grade in the assignment.
Appendix A
Instruction Set Architecture (ISA)
There are 13 arithmetic/logical operations that manipulate the data within data-stack. These operations are indicated by the OP component = 2 (OPR). When an OPR instruction is encountered, the M component of the instruction is used to select the arithmetic/logical operation to be executed (e.g. to multiply the two elements at the top of the stack, write the instruction 2 0 4).
ISA:
01 LIT 0, M Push constant value (literal) M onto the stack
02 OPR 0, M Operation to be performed on the data at the top of the stack
(See in Appendix B)
03 LOD L, M Load value to top of stack from the stack location at offset M from
L lexicographical levels down
04 STO L, M Store value at top of stack in the stack location at offset M from
L lexicographical levels down
05 CAL L, M Call procedure at code index M (generates new Activation Record
and pc M)
06 INC 0, M Allocate M locals (increment sp by M). First three are Static Link
(SL), Dynamic Link (DL), and Return Address (RA)
07 JMP 0, M Jump to instruction M
08 JPC 0, M Jump to instruction M if top stack element is 0
09 SIO 0, 1 Write the top stack element to the screen
10 SIO 0, 2 Read in input from the user and store it on top of the stack.
11 SIO 0, 3 End of Program.
Appendix B
ISA Pseudo Code
01 LIT 0, M sp sp 1;
stack[sp] M;
02 OPR 0, # ( 0 # 13 )
0 RET sp bp + 1
pc stack[sp 4]
bp stack[sp 3])
1 NEG (-stack[sp])2 ADD (sp sp + 1; stack[sp] stack[sp] + stack[sp 1 ])3 SUB (sp sp + 1; stack[sp] stack[sp] stack[sp 1])
4 MUL (sp sp + 1; stack[sp] stack[sp] * stack[sp 1])
5 DIV (sp sp + 1; stack[sp] stack[sp] / stack[sp 1])6 ODD (stack[sp] stack[sp] mod 2) or ord(odd(stack[sp]))7 MOD (sp sp + 1; stack[sp] stack[sp] mod stack[sp 1])8 EQL (sp sp + 1; stack[sp] stack[sp] = = stack[sp 1])9 NEQ (sp sp + 1 and stack[sp] stack[sp] != stack[sp 1])10 LSS (sp sp + 1 and stack[sp] stack[sp] < stack[sp 1])
11 LEQ (sp sp + 1 and stack[sp] stack[sp] <= stack[sp 1])12 GTR (sp sp + 1 and stack[sp] stack[sp] > stack[sp 1])
13 GEQ (sp sp + 1 and stack[sp] stack[sp] >= stack[sp 1])
03 LOD L, M sp sp 1;
stack[sp] stack[ base(L, bp) M];
04 STO L, M stack[ base(L, bp) M] stack[sp];
sp sp + 1;
05 CAL L, M stack[sp 1] 0 /* space to return value
stack[sp 2] base(L, bp); /* static link (SL)
stack[sp 3] bp; /* dynamic link (DL)
stack[sp 4] pc /* return address (RA)
bp sp 1;
pc M;
06 INC 0, M sp sp M;
07 JMP 0, M pc M;
08 JPC 0, M if stack[sp] == 0 then { pc M; }
sp sp + 1;
09 SIO 0, 1 print(stack[sp]);
sp sp + 1;
10 SIO 0, 2 sp sp 1;
read(stack[sp]);
11 SIO 0, 3 Set Halt flag to zero;
NOTE: The result of a logical operation such as (A > B) is defined as 1 ifthe condition was met and 0 otherwise.
Appendix C
Example of Execution
This example shows how to print the stack after the execution of each instruction. The following PL/0 program, once compiled, will be translated into a sequence code for the virtual machine PM/0 as shown below in the INPUT FILE.
const n = 9;
int i,h;
procedure sub;
const k = 7;
int j,h;
begin
j:=n;
i:=1;
h:=k;
end;
begin
i:=3; h:=8;
call sub;
end.
INPUT FILE
For every line, there must be 3 integers representing OP, L and M.
7 0 10
7 0 2
6 0 6 we recommend using the following structure for your instructions:
1 0 9
4 0 4 struct {
1 0 1 int op; /* opcode
4 1 4 int l; /* L
1 0 7 int m; /* M
4 0 5 }instruction;
2 0 0
6 0 6
1 0 3
4 0 4
1 0 8
4 0 5
5 0 2
11 0 3
OUTPUT FILE
1) Print out the program in interpreted assembly language with line numbers:
Line OP L M
0 jmp 0 10
1 jmp 0 2
2 inc 0 6
3 lit 0 9
4 sto 0 4
5 lit 0 1
6 sto 1 4
7 lit 0 7
8 sto 0 5
9 opr 0 0
10 inc 0 6
11 lit 0 3
12 sto 0 4
13 lit 0 8
14 sto 0 5
15 cal 0 2
16 sio 0 3
2) Print out the execution of the program in the virtual machine, showing the stack and pc, bp, and sp:
pc bp sp stack
Initial values 0 999 1000
0 jmp 0 10 10 999 1000
10 inc 0 6 11 999 994 0 0 0 0 0 0
11 lit 0 3 12 999 993 0 0 0 0 0 0 3
12 sto 0 4 13 999 994 0 0 0 0 3 0
13 lit 0 8 14 999 993 0 0 0 0 3 0 8
14 sto 0 5 15 999 994 0 0 0 0 3 8
15 cal 0 2 2 993 994 0 0 0 0 3 8
2 inc 0 6 3 993 988 0 0 0 0 3 8 | 0 999 999 16 0 0
3 lit 0 9 4 993 987 0 0 0 0 3 8 | 0 999 999 16 0 0 9
4 sto 0 4 5 993 988 0 0 0 0 3 8 | 0 999 999 16 9 0
5 lit 0 1 6 993 987 0 0 0 0 3 8 | 0 999 999 16 9 0 1
6 sto 1 4 7 993 988 0 0 0 0 1 8 | 0 999 999 16 9 0
7 lit 0 7 8 993 987 0 0 0 0 1 8 | 0 999 999 16 9 0 7
8 sto 0 5 9 993 988 0 0 0 0 1 8 | 0 999 999 16 9 7
9 opr 0 0 16 999 994 0 0 0 0 1 8
16 sio 0 3 17 999 994
NOTE: It is necessary to separate each Activation Record with a bracket |.
Appendix D
Helpful Tips
This function will be helpful to find a variable in a different Activation Record some L levels down:
/**********************************************/
/* Find base L levels down */
/* */
/**********************************************/
int base(l, base) // l stand for L in the instruction format{ int b1; //find base L levels down b1 = base; while (l > 0) { b1 = stack[b1 1]; l; } return b1;}
For example in the instruction:
STO L, M you can do stack[base(ir[pc].L, bp) + ir[pc].M] = stack[sp] to store a variable L levels down.
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