### Input SpecificationsYour simulator must run from the command line with a single input file as a parameter to main. This file will contain a sequence of instructions for your simulator to store in "Instruction Memory" and then run via the fetch/execute cycle. In the input file, each instruction is represented with two integers: the first one represents the opcode and the second one a memory address or a device number depending on the instruction.#### Example:**Input File**```5 5 // IN 56 7 // OUT 73 0 // STORE 05 5 // IN 56 7 // OUT 73 1 // STORE 11 0 // LOAD 04 1 // SUB 13 0 // STORE 06 7 // OUT 71 1 // LOAD 16 7 // OUT 77 0 // END```### Output SpecificationsYour simulator should provide output according to the input file. Along with this output, your program should provide status messages identifying details on the workings of your simulator. Output text does not have to reflect the example word-for-word, but please provide detail on the program as it runs in a readable format that does not conflict with the actual output of your simulator. After each instruction, print the current state of the Program Counter, Accumulator, and Data Memory. The INPUT instruction is the only one that should prompt an interaction from the user. **The Problem**Using the C programming language, write a program that simulates a variant of the Tiny Machine Architecture. In this implementation, memory (RAM) is split into Instruction Memory (IM) and Data Memory (DM). Your code must implement the basic instruction set architecture (ISA) of the Tiny Machine Architecture:1. 1 → LOAD2. 2 → ADD3. 3 → STORE4. 4 → SUB5. 5 → IN6. 6 → OUT7. 7 → END8. 8 → JMP9. 9 → SKIPZEach piece of the architecture must be accurately represented in your code (Instruction Register, Program Counter, Memory Address Registers, Instruction Memory, Data Memory, Memory Data Registers, and Accumulator). Data Memory will be represented by an integer array. Your Program Counter will begin pointing to the first instruction of the program.**For the sake of simplicity, Instruction Memory (IM) and Data Memory (DM) may be implemented as separate arrays.****Hint:** Implementing a struct for your Instructions and an array of these structs as your Instruction Memory greatly simplifies this program.**Example:**```ctypedef struct { int opCode, deviceOrAddress;} Instruction;Instruction IM[MAXPROGRAMSIZE];```**Note:** IM, MDR1, and IR are of type Instruction. All other CPU registers and Data Memory (DM) are of type int.

PROGRAM USING C : #include <stdio.h>#include <stdlib.h>#include <string.h> int hexnum2int( char* hexnum){int result=0;while ((*hexnum)!='\0'){if (('0'<=(*hexnum))&&((*hexnum)<='9'))result = result*16 + (*hexnum) -'0';else if (('a'<=(*hexnum))&&((*hexnum)<='f'))result = result*16 + (*hexnum) -'a'+10;else if (('A'<=(*hexnum))&&((*hexnum)<='F'))result = result*16 + (*hexnum) -'A'+10;hexnum++;}return(result);} void main( int argc, char *argv[]){FILE *f;char line[100];char *token = NULL;char *option1, *option2, *option3, *label;char ch;int ch1;int inst[1000];int count=0;struct label{int loc;char *label;};struct label labeltab[50];int numlabels = 0;struct jumpinst{int loc;char *label;};struct jumpinst jumptab[100];int numjumps=0; struct var{int loc;char *name;};struct var vartab[50];int numvars = 0; struct ldiinst{int loc;char *name;};struct ldiinst lditab[100];int numldis=0; f = fopen(argv[1],"r");if (f != NULL){while(fgets(line,sizeof line,f)!= NULL){token=strtoken(line,"\n\t\r ");if (strcmp(token,".code")==0 )break;}while(fgets(line,sizeof line,f)!= NULL){token=strtoken(line,"\n\t\r ");while (token){if (strcmp(token,"ldi")==0){option1 = strtoken(NULL,"\n\t\r ");option2 = strtoken(NULL,"\n\t\r ");inst[count]=0x1000+hexnum2int(option1);count++; if ((option2[0]=='0')&&(option2[1]=='x'))inst[count]=hexnum2int(option2+2)&0xffff;else if (( (option2[0])=='-') || ((option2[0]>='0')&&(option2[0]<='9')))inst[count]=atoi(option2)&0xffff;else { lditab[numldis].loc = count; option1=(char*)malloc(sizeof(option2));strcpy(option1,option2); lditab[numldis].name = option1;numldis++;}count++; }else if (strcmp(token,"ld")==0){option1 = strtoken(NULL,"\n\t\r ");option2 = strtoken(NULL,"\n\t\r ");ch = (option1[0]-48)| ((option2[0]-48) << 3);inst[count]=0x2000+((ch)&0x00ff);count++; }else if (strcmp(token,"st")==0){}else if (strcmp(token,"jz")==0){}else if (strcmp(token,"jmp")==0){option1 = strtoken(NULL,"\n\t\r ");jumptab[numjumps].loc = count;option2=(char*)malloc(sizeof(option1));strcpy(option2,option1);jumptab[numjumps].label=option2;numjumps++; inst[count]=0x5000; count++;}else if (strcmp(token,"add")==0){option1 = strtoken(NULL,"\n\t\r "); option2 = strtoken(NULL,"\n\t\r ");option3 = strtoken(NULL,"\n\t\r ");ch1 = (option1[0]-48)| ((option2[0]-48)<<3)|((option3[0]-48)<<6);inst[count]=0x7000+((ch1)&0x00ff);count++;}else if (strcmp(token,"sub")==0){}else if (strcmp(token,"and")==0){}else if (strcmp(token,"or")==0){}else if (strcmp(token,"xor")==0){}else if (strcmp(token,"not")==0){option1 = strtoken(NULL,"\n\t\r ");option2 = strtoken(NULL,"\n\t\r ");ch = (option1[0]-48)| ((option2[0]-48)<<3);inst[count]=0x7500+((ch)&0x00ff);count++;}else if (strcmp(token,"mov")==0){}else if (strcmp(token,"inc")==0){option1 = strtoken(NULL,"\n\t\r ");ch = (option1[0]-48)| ((option1[0]-48)<<3);inst[count]=0x7700+((ch)&0x00ff);count++;}else if (strcmp(token,"dec")==0){}else{labeltab[numlabels].loc = count;option1=(char*)malloc(sizeof(token));strcpy(option1,token);labeltab[numlabels].label=option1;numlabels++;}token = strtoken(NULL,",\n\t\r ");}} int i,j;for (i=0; i<numjumps;i++){j=0;while ( strcmp(jumptab[i].label , labeltab[j].label) != 0 )j++;inst[jumptab[i].loc] +=(labeltab[j].loc-jumptab[i].loc-1)&0x0fff;} rewind(f);while(fgets(line,sizeof line,f)!= NULL){token=strtoken(line,"\n\t\r ");if (strcmp(token,".data")==0 )break;} int data_area=0;while(fgets(line,sizeof line,f)!= NULL){token=strtoken(line,"\n\t\r ");if (strcmp(token,".code")==0 )break;else if (token[strlen(token)-1]==':'){token[strlen(token)-1]='\0';vartab[numvars].loc=count+data_area;option1=(char*)malloc(sizeof(token));strcpy(option1,token);vartab[numvars].name=option1;token = strtoken(NULL,",\n\t\r ");if (token==NULL)inst[count+data_area]=0;else if (strcmp(token, ".space")==0){token=strtoken(NULL,"\n\t\r ");data_area+=atoi(token);}else if((token[0]=='0')&&(token[1]=='x'))inst[count+data_area]=hexnum2int(token+2)&0xffff;else if (( (token[0])=='-') || ('0'<=(token[0])&&(token[0]<='9')) )inst[count+data_area]=atoi(token)&0xffff;numvars++;data_area++;}} for( i=0; i<numldis;i++){j=0;while ((j<numvars)&&( strcmp( lditab[i].name , vartab[j].name)!=0 ))j++;if (j<numvars)inst[lditab[i].loc] = vartab[j].loc; } for( i=0; i<numldis;i++){j=0;while ((j<numlabels)&&( strcmp( lditab[i].name , labeltab[j].label)!=0 ))j++;if (j<numlabels){inst[lditab[i].loc] = (labeltab[j].loc)&0x0fff;printf("%d %d %d\n", i, j, (labeltab[j].loc));}} printf("LABEL tab\n");for (i=0;i<numlabels;i++)printf("%d %s\n", labeltab[i].loc, labeltab[i].label);printf("\n");printf("JUMP tab\n");for (i=0;i<numjumps;i++)printf("%d %s\n", jumptab[i].loc, jumptab[i].label);printf("\n");printf("var tab\n");for (i=0;i<numvars;i++)printf("%d %s\n", vartab[i].loc, vartab[i].name); printf("\n");printf("LDI instS\n");for (i=0;i<numldis;i++)printf("%d %s\n", lditab[i].loc, lditab[i].name);printf("\n");fclose(f);f = fopen("RAM","w");frintf(f,"v2.0 raw\n");for (i=0;i<count+data_area;i++)frintf(f,"%04x\n",inst[i]);}}

Engineering

Computer Engineering

Your simulator must run from the command line with a single input file as a parameter to main. This file will contain a sequence of instructions for your simulator to store in "Instruction Memory" and then run via the fetch/execute cycle. In the input file each instruction is represented with two integers: the first one represents the opcode and the second one a memory address or a device number depending on the instruction.

Your simulator must run from the command line with a single input file as a parameter to main. This file will contain a sequence of instructions for your simulator to store in "Instruction Memory" and then run via the fetch/execute cycle. In the input file each instruction is represented with two integers: the first one represents the opcode and the second one a memory address or a device number depending on the instruction.

Computer Networking: A Top-Down Approach (7th Edition)

7th Edition

ISBN:9780133594140

Author:James Kurose, Keith Ross

Publisher:James Kurose, Keith Ross

Chapter1: Computer Networks And The Internet

Section: Chapter Questions

Problem R1RQ: What is the difference between a host and an end system? List several different types of end...

See similar textbooks

Similar questions

Does a computer execute the memory fetch operation right away when it requires a piece of information? What does it do if not that?
Task 'A' run on a CPU. The task starts by running for 6 ms. 'A' waits for 4 ms while the operating system runs some instructions to access memory. The CPU is then idle for 1 ms while waiting for the memory read to complete. Finally, the task 'A' runs another 25 ms and completes. What is the CPU performance in ms? [ Give answer as a number only]
TRUE OR FALSE: The instruction "inc [esi]" is valid and won't throw an error.
This article elaborates upon the concept of memory hierarchy. For its visual appeal, your writers chose a pyramid for a variety of reasons.
It's a counter for both the number of source-level statements and the total number of I/O operations.
There has to be a rundown of the four memory allocation methods covered in Operating Systems. Just how often do you utilize the other two?
ASSEMBLY There is one instruction missing that will make this code display the right results. It will compile in its current state, but you will not get what you expect. What is that instruction? (Hint: It is where the square brackets are indicated) .datavar1 DWORD 4 var2 DWORD 2 .code [code] mov eax, var1 mov ebx, var2 div ebx
Does the memory fetch procedure always happen instantly when a computer wants data? So what is the point of it?
What is the difference between a debug command and a CPU instruction?
Whenever the CPU receives a signal, it stops whatever it is doing to deal with the incoming request. So, here's how the mechanism operates: Spooling with B as the interrupt signal C as the interrupt handler D as the polling method A'.
With virtual memory, when a program requests memory,the memory management system does which of the following? Select one: A. It grants the request as long as there is enough physical memory to accommodate B. It looks up the address in the page table. If the address is in memory, then it grants the request. C. It looks up the address in the page table. If the address is in memory, it grants the request. If the address is stored to disk, it writes a section of memory to disk and then transfers the requested virtual memory address from disk to main memory. D. It causes a deadlock to occur.
What's the difference between a memory-resident program and a handler for interrupts?