SMACO Simulator

Assignment 2 : SMAC0 simulator

 Software Description – A simple instruction Computer(SMAC0) is a hypothetical machine with a small but effective instruction set that can be used to illustrate the design of simple software processors involved in development of programs such as Assembler, Macro processor etc.

 The machine will incorporate the most commonly encountered hardware features and concepts, while avoiding irrelevant complexities. 

 A simulator program is required that simulates the function of simple instruction computer such as fetching an instruction, decoding and executing it. 

The hypothetical Simple machine (SMAC0) has following features.

#  Memory – Memory consist of 6 digit words (decimal). 

#  Total size of memory is 1000 words (103 ) indicates the address size is 3 digits ( address ranges from 0 to 999). 

# Registers – There are in all six registers four general purpose registers AREG, BREG, CREG and DREG numbered 1,2,3 and 4.

#  A program counter (PC) storing the address of the next instruction to be fetched and a status register storing condition codes. 

# There are SIX condition codes LT, LE, EQ, GT , GE and ANY numbered 0,1,2,3,4 and 5. Each bit in the status register can be set to 1 

# Data Format – Supports only six digit integer data stored in decimal form.

#  Instruction Format – It has single instruction format. Each instruction is of six digit length. The opcode, register operand and memory operand occupy 2, 1 and 3 digits in that order.

Instruction Set

 Opcode  Mnemonic Instruction   Operands 

00 ---STOP ----Stop or Halt execution Operands unused

 01 ----ADD ----Add memory operand to register operand Register and memory operand

 02---- SUB---- Subtract memory operand from register operand Register and memory operand 

03---- MULT---- Multiply memory operand to register operand Register and memory operand 

08---- DIV ----Divide register operand by memory operand Register and memory operand 

04---- MOVER ----Move memory operand contents to register operand Register and memory operand 

05---- MOVEM---- Move register operand contents to memory Register and memory operand

 06 ----COMP---- Compare register and memory operands to set condition code appropriately Register and memory operand 

07---- BC---- Branch to second operand depending on condition code specified as first operand Register and memory operand 

09 ----READ ----Read into memory operand Only memory operand 

10 ----PRINT ----Print contents of memory Operand Only memory operand 

Condition code Mnemonic Description 

0 ---LT---- Less than 

1--- LE ----Less than or equal to 

2--- EQ ---Equal to 

3 ---GT--- Greater than 

4--- GE ---Greater than or equal to

5---- ANY ----Unconditional 

Slot 1

i) Answer the following questions after carefully reading the description and program structure. 

a) What is the size of memory in Hypothetical Simple Instruction Computer(SMAC0)? How memory will be represented in simulation program?

Answer:- 

Memory – Memory consist of 6 digit words (decimal). Total size of memory is 1000 words (103 ) indicates the address size is 3 digits ( address ranges from 0 to 999).

b) How many registers are there in SMAC0? How registers are represented in simulation program? 

Answer:- 

Registers – There are in all six registers four general purpose registers AREG, BREG, CREG and DREG numbered 1,2,3 and 4. 

A program counter (PC) storing the address of the next instruction to be fetched and a status register storing condition codes. There are SIX condition codes LT, LE, EQ, GT , GE and ANY numbered 0,1,2,3,4 and 5. Each bit in the status register can be set to 1 

c) From the contents of the memory at pc, how will you separate opcode, register operand and memory operand? 

Answer:- 

mem[pc] // contains the instruction 

opcode=mem[pc]/10000; // separate register and memory operand

 ii) Implement a menu driven simulator for hypothetical Simple Instruction Computer that provides the following functionalities 

a) Load - Loading of the program from file into memory 

b) Print - Printing the program loaded in memory 

c) Run - Executing the loaded program

 The machine has the basic instruction set comprising of add, mover, movem, read, print and hlt commands as given in Table 1. Create a file sum.sm containing the machine code for sum of two numbers. Test the program using the machine code programs sum.sm.

Program :- 

#include<stdio.h>

#include<stdlib.h>

int mem[100],reg[4],cond[]={0,0,0,0,0,1},opc,op1,op2,pc;

FILE *fp;

char fname[20];

void load()

{

printf("Enter file name:");

scanf("%s",fname);

fp = fopen(fname,"r");

if(fp==NULL)

{

printf("File %s not found.\n",fname);

exit(1);

}

while(fscanf(fp,"%d",&mem[pc])!=-1)

pc++;

fclose(fp);

}

void print()

{

int i;

for(i=0;i<pc;i++)

printf("%06d\n",mem[i]);

}

void run()

{ int i;

pc = 0;

while(1)

{ opc = mem[pc]/10000;

op1 = mem[pc]%10000/1000-1;

op2 = mem[pc]%1000;

switch(opc)

{

case 0: // STOP

return;

case 1: // ADD

reg[op1]+=mem[op2];

break;

case 2: // SUB

reg[op1]-=mem[op2];

break;

case 3: // MULT

reg[op1]*=mem[op2];

break;

case 4: // DIV

reg[op1]/=mem[op2];

break;

case 5: // MOVER

reg[op1]=mem[op2];

break;

case 6: // MOVEM

mem[op2]=reg[op1];

break;

case 7: // COMP

if(reg[op1] < mem[op2])

cond[0]=1;

if(reg[op1] <= mem[op2])

cond[1]=1;

if(reg[op1] == mem[op2])

cond[2]=1;

if(reg[op1] > mem[op2])

cond[3]=1;

if(reg[op1] >= mem[op2])

cond[4]=1;

break;

case 8: // BC

if(cond[op1]==1)

pc = op2-1;

for(i=0;i<5;i++)

cond[i]=0;

break;

case 9: // READ

scanf("%d",&mem[op2]);

break;

case 10:// PRINT

printf("%d\n",mem[op2]);

}

pc++;

}

}

int main()

{ int ch;

while(1)

{

printf("1.Load\n2.Print\n3.Run\n4.Exit\n");

printf("Enter your choice (1-4):");

scanf("%d",&ch);

switch(ch)

{

case 1:

load();

break;

case 2:

print();

break;

case 3:

run();

break;

case 4:

exit(0);

}

}

return 0;

}

-------------------------------------------------------------------------------------------

OUTPUT:-

1.Load

2.Print

3.Run

4.Exit

Enter your choice (1-4):1

Enter file name:sum.x

1.Load

2.Print

3.Run

4.Exit

Enter your choice (1-4):2

090014

051017

061016

061015

071014

084012

052016

012015

062016

051015

011018

086003

100016

000000

000000

000000

000000

000000

000001

1.Load

2.Print

3.Run

4.Exit

Enter your choice (1-4):3

1

2

3

1.Load

2.Print

3.Run

4.Exit

Enter your choice (1-4):4

------------------------------------------------------------------

sum.x

090014

051017

061016

061015

071014

084012

052016

012015

062016

051015

011018

086003

100016

000000

0

0

0

0

1