Help with creating a virtual machine. I dont understand how to print the stack trace.

This is the input:

6 0 0 6 1 0 0 3 4 0 0 4 1 0 0 1 4 0 0 5 5 0 0 7 7 0 0 19 6 0 0 4 3 0 1 4 3 1 1 5 13 1 0 1 4 1 1 5 1 1 0 1 12 0 0 1 4 0 1 4 18 0 0 1 8 0 0 18 5 0 1 7 2 0 0 0 3 0 0 5 9 0 0 1 9 0 0 3

This is the output:

Factorial Op Printout:

0 inc 0 0 6

1 lit 0 0 3

2 sto 0 0 4

3 lit 0 0 1

4 sto 0 0 5

5 cal 0 0 7

6 jmp 0 0 19

7 inc 0 0 4

8 lod 0 1 4

9 lod 1 1 5

10 mul 1 0 1

11 sto 1 1 5

12 lit 1 0 1

13 sub 0 0 1

14 sto 0 1 4

15 neq 0 0 1

16 jpc 0 0 18

17 cal 0 1 7

18 rtn 0 0 0

19 lod 0 0 5

20 sio 0 0 1

21 sio 0 0 3

Factorial Stack Trace:

Inital Values pc bp sp

0 inc 0 0 6 1 1 6 0 1 0 0 0 0

1 lit 0 0 3 2 1 6 0 1 0 0 0 0

2 sto 0 0 4 3 1 6 0 1 0 0 3 0

3 lit 0 0 1 4 1 6 0 1 0 0 3 0

4 sto 0 0 5 5 1 6 0 1 0 0 3 1

5 cal 0 0 7 7 7 6 0 1 0 0 3 1

7 inc 0 0 4 8 7 10 0 1 0 0 3 1 | 0 1 1 6

8 lod 0 1 4 9 7 10 0 1 0 0 3 1 | 0 1 1 6

9 lod 1 1 5 10 7 10 0 1 0 0 3 1 | 0 1 1 6

10 mul 1 0 1 11 7 10 0 1 0 0 3 1 | 0 1 1 6

11 sto 1 1 5 12 7 10 0 1 0 0 3 3 | 0 1 1 6

12 lit 1 0 1 13 7 10 0 1 0 0 3 3 | 0 1 1 6

13 sub 0 0 1 14 7 10 0 1 0 0 3 3 | 0 1 1 6

14 sto 0 1 4 15 7 10 0 1 0 0 2 3 | 0 1 1 6

15 neq 0 0 1 16 7 10 0 1 0 0 2 3 | 0 1 1 6

16 jpc 0 0 18 17 7 10 0 1 0 0 2 3 | 0 1 1 6

17 cal 0 1 7 7 11 10 0 1 0 0 2 3 | 0 1 1 6

7 inc 0 0 4 8 11 14 0 1 0 0 2 3 | 0 1 1 6 | 0 1 7 18

8 lod 0 1 4 9 11 14 0 1 0 0 2 3 | 0 1 1 6 | 0 1 7 18

9 lod 1 1 5 10 11 14 0 1 0 0 2 3 | 0 1 1 6 | 0 1 7 18

10 mul 1 0 1 11 11 14 0 1 0 0 2 3 | 0 1 1 6 | 0 1 7 18

11 sto 1 1 5 12 11 14 0 1 0 0 2 6 | 0 1 1 6 | 0 1 7 18

12 lit 1 0 1 13 11 14 0 1 0 0 2 6 | 0 1 1 6 | 0 1 7 18

13 sub 0 0 1 14 11 14 0 1 0 0 2 6 | 0 1 1 6 | 0 1 7 18

14 sto 0 1 4 15 11 14 0 1 0 0 1 6 | 0 1 1 6 | 0 1 7 18

15 neq 0 0 1 16 11 14 0 1 0 0 1 6 | 0 1 1 6 | 0 1 7 18

16 jpc 0 0 18 18 11 14 0 1 0 0 1 6 | 0 1 1 6 | 0 1 7 18

18 rtn 0 0 0 18 7 10 0 1 0 0 1 6 | 0 1 1 6

18 rtn 0 0 0 6 1 6 0 1 0 0 1 6

6 jmp 0 0 19 19 1 6 0 1 0 0 1 6

19 lod 0 0 5 20 1 6 0 1 0 0 1 6

20 sio 0 0 1 21 1 6 0 1 0 0 1 6

21 sio 0 0 3 0 1 0 0

Factorial Output:

6

Here are the operation instructions:

The code (curious to know if I have an error for case 9):

#include

#include

#define MAX_STACK_HEIGHT 2000

#define MAX_CODE_LENGTH 500

#define MAX_LEXI_LEVELS 3

int base(int l, int base);

void fetch(void);

void execute(void);

void print(int count);

void printStack(void);

typedef struct instruction

{

int op; //opcode

int r; // reg

int l; // L

int m; // M

} instruction;

int stack[MAX_STACK_HEIGHT] = {0};

int sp = 0, bp = 1, pc = 0;

int r[16] = {0};

instruction code[MAX_CODE_LENGTH];

instruction ir;

FILE *ifp;

int halt = 0;

int main(void)

{

ifp = fopen("code.txt", "r");

if (ifp == NULL)

{

printf("File could not open. Program halted. ");

}

//Keep track of number of instructions the input file.

int count = 0;

//Read in the lines from the input file.

while(!feof(ifp))

{

fscanf(ifp, "%d", &code[count].op);

fscanf(ifp, "%d", &code[count].r);

fscanf(ifp, "%d", &code[count].l);

fscanf(ifp, "%d", &code[count].m);

count++;

}

fprintf(ifp, "%d", code[count].op);

print(count);

count = 0;

printf(" Factorial Stack Trace: ");

printf("Initial Values\t\tpc\tbp\tsp ");

while (halt == 0)

{

//Start the fetch part of the fetch-execute cycle.

fetch();

//Start the execution process of the

execute();

printf("%d\t%d\t%d\t", ir.r, ir.l, ir.m);

printf("%d\t%d\t%d\t", pc, bp, sp);

printStack();

printf(" ");

if ((pc == 0) && (bp == 0) && (sp == 0))

halt = 1;

}

printf(" Factorial Output: %d ", r[ir.r]);

fclose(ifp);

return 0;

}

void print(int count)

{

int i;

printf("Factorial OP Printout:");

for (i = 0; i

{

int op = code[i].op;

//Interpret the operation.

switch(op)

{

case 1: printf(" %d\t", i); printf("lit\t"); break;

case 2: printf(" %d\t", i); printf("rtn\t"); break;

case 3: printf(" %d\t", i); printf("lod\t"); break;

case 4: printf(" %d\t", i); printf("sto\t"); break;

case 5: printf(" %d\t", i); printf("cal\t"); break;

case 6: printf(" %d\t", i); printf("inc\t"); break;

case 7: printf(" %d\t", i); printf("jmp\t"); break;

case 8: printf(" %d\t", i); printf("jpc\t"); break;

case 9: printf(" %d\t", i); printf("sio\t"); break;

case 10:printf(" %d\t", i); printf("sio\t"); break;

case 11:printf(" %d\t", i); printf("sio\t"); break;

case 12:printf(" %d\t", i); printf("neg\t"); break;

case 13:printf(" %d\t", i); printf("add\t"); break;

case 14:printf(" %d\t", i); printf("sub\t"); break;

case 15:printf(" %d\t", i); printf("mul\t"); break;

case 16:printf(" %d\t", i); printf("div\t"); break;

case 17:printf(" %d\t", i); printf("odd\t"); break;

case 18:printf(" %d\t", i); printf("mod\t"); break;

case 19:printf(" %d\t", i); printf("eql\t"); break;

case 20:printf(" %d\t", i); printf("neq\t"); break;

case 21:printf(" %d\t", i); printf("lss\t"); break;

case 22:printf(" %d\t", i); printf("leq\t"); break;

}

printf("%d\t", code[i].r);

printf("%d\t", code[i].l);

printf("%d\t", code[i].m);

}

return;

}

void fetch()

{

ir = code[pc];

pc++;

return;

}

void execute()

{

switch(ir.op)

{

//LIT

case 1:

printf("lit\t");

r[ir.r] = ir.m;

break;

//RTN

case 2:

printf("rtn\t");

sp = bp - 1;

bp = stack[sp+3];

pc = stack[sp+4];

break;

//LOD

case 3:

printf("lod\t");

r[ir.r] = stack[base(ir.l, bp) + ir.m];

break;

//STO

case 4:

printf("sto\t");

stack[base(ir.l, bp) + ir.m] = r[ir.r];

break;

//CAL

case 5:

printf("cal\t");

stack[sp + 1] = 0;

stack[sp + 2] = base(ir.l, bp);

stack[sp + 3] = bp;

stack[sp + 4] = pc;

bp = sp + 1;

pc = ir.m;

break;

//INC

case 6:

printf("inc\t");

sp = sp + ir.m;

break;

//JMP

case 7:

printf("jmp\t");

pc = ir.m;

break;

//JPC

case 8:

printf("jpc\t");

if (r[ir.r] == 0)

pc = ir.m;

break;

//SIO

case 9:

printf("sio\t");

if(ir.m == 1)

printf(" R[%d] = %d ", ir.r, r[ir.r]);

else if (ir.m == 2)

{

printf(" Enter number to store in register. ");

scanf("%d", &r[ir.r]);

}

else if (ir.m == 3)

halt = 1;

break;

//NEG

case 10:

printf("neg\t");

r[ir.r] = 0 - r[ir.l];

break;

//ADD

case 11:

printf("add\t");

r[ir.r] = r[ir.l] + r[ir.m];

break;

//SUB

case 12:

printf("sub\t");

r[ir.r] = r[ir.l] - r[ir.m];

break;

//MUL

case 13:

printf("mul\t");

r[ir.r] = r[ir.l] * r[ir.m];

break;

//DIV

case 14:

printf("div\t");

r[ir.r] = r[ir.l] / (r[ir.m]);

break;

//ODD

case 15:

printf("odd\t");

r[ir.r] = r[ir.r] % 2;

break;

//MOD

case 16:

printf("mod\t");

r[ir.r] = r[ir.l] % r[ir.m];

break;

//EQL

case 17:

printf("eql\t");

if (r[ir.l] == r[ir.m])

r[ir.r] = 1;

else

r[ir.r] = 0;

break;

//NEQ

case 18:

printf("neq\t");

if (r[ir.l] != r[ir.m])

r[ir.r] = 1;

else

r[ir.r] = 0;

break;

//LSS

case 19:

printf("lss\t");

if (r[ir.l]

r[ir.r] = 1;

else

r[ir.r] = 0;

break;

//LEQ

case 20:

printf("leq\t");

if (r[ir.l]

r[ir.r] = 1;

else

r[ir.r] = 0;

break;

//GTR

case 21:

printf("gtr\t");

if (r[ir.l] > r[ir.m])

r[ir.r] = 1;

else

r[ir.r] = 0;

break;

//GEQ

case 22:

printf("geq\t");

if (r[ir.l] >= r[ir.m])

r[ir.r] = 1;

else

r[ir.r] = 0;

break;

default:

printf("err\t");

return;

}

return;

}

void printStack()

{

if (bp == 0)

return;

else

{

int i;

for(i = 0; i

printf("%d\t", stack[i]);

return;

}

}

int base(int l, int 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;

}

Appendix B ISA Pseudo Cod