Operating Systems COMP 310 – ECSE 427 McGill University Vybihal Assignment #2 Page 1 of 6 Assignment #2 The Kernel and Process Execution Due: February 26, 2019 at 23:55 on myCourses Labs 3 and 4 will...

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Operating Systems COMP 310 – ECSE 427 McGill University Vybihal Assignment #2 Page 1 of 6 Assignment #2 The Kernel and Process Execution Due: February 26, 2019 at 23:55 on myCourses Labs 3 and 4 will provide some background for this assignment. The question is presented from a Linux point of view using the computer science server mimi.cs.mcgill.ca, which you can reach remotely using ssh or putty from your laptop (see lab 1). If you do this assignment from an MS Windows machine, then make sure to provide the DLL libraries your program uses (if any) so that the TA can run it from their MS Windows machine. It is not the TA’s responsibility to make your program run. The TAs will not debug your program. You must write this assignment in the C Programming language. Assignment Question: Building a Kernel This assignment builds from assignment 1. You will add a new command line command: exec prog1 prog2 prog3 This new command will simulate the kernel run-time environment, which includes the PCB, the ready queue, the CPU, and a simple memory. To make this easier we will repurpose code from assignment 1. In assignment 2 you will maintain a fully working assignment 1 and add to it a kernel run time environment. You will need to implement the following data structures: CPU, PCB, ready queue, and RAM. You will need to implement the following algorithms: schedular, task switch, and basic memory management. You are NOT implementing threading. This assignment first provides a description of the files and data structures and how they work. Then a description of the new shell command and how it works. Lastly a description of the new algorithms. Your source files: Your entire application must contain the source files you used from assignment 1: shell.c, interpreter.c, and shellmemory.c. You must add the following assignment 2 files: kernel.c, cpu.c, pcb.c, and ram.c. Add .h files as you see fit. These source files must be built using modular programming techniques (see lab 2). Move the main() function from shell.c to kernel.c. This is the true home of the main() function. The kernel.c main() function calls the shell.c UI function (maybe your previous main() function), to display the command-line to the user. The kernel main() will also instantiate all the kernel data structures. Compiling assignment 2: Compile your application using gcc with the name mykernel. To do modular programming, you must compile your application in the following manner: gcc -c shell.c interpreter.c shellmemory.c kernel.c cpu.c pcb.c ram.c gcc -o mysh shell.o interpreter.o shellmemory.o kernel.o cpu.o pcb.o ram.o Operating Systems COMP 310 – ECSE 427 McGill University Vybihal Assignment #2 Page 2 of 6 Running assignment 2: From the command line prompt type: ./mykernel What mykernel displays to the user: Kernel 1.0 loaded! Welcome to the shell! Shell version 2.0 Updated February 2019 $ The above dollar sign is the prompt. The cursor flashes beside the prompt waiting for the user’s input. From this prompt the user will type in their command and the shell will display the result from that command on the screen (or an error message), and then the dollar sign is displayed again prompting the next command. The user stays in this interface until they ask to quit. Your shell from assignment #1 is fully functional in this assignment. A single new command is added to the shell, called exec. New command supported by your shell: COMMAND DESCRIPTION exec p1 p2 p3 Executes concurrent programs $ exec prog.txt prog2.txt The exec command takes one to three arguments. Each argument is the name of a different mysh script filename. For this assignment, exec does not permit us to launch multiple scripts with the same filename. If you try to do that your shell displays the error, “Error: Script already loaded”. It will however continue to load/exec the other scripts. To simplify the assignment, we will assume that “compiled” programs are the same as mysh scripts. We will reuse the shell’s interpreter to both run shell scripts and execute the kernel programs. A more detailed description follows. Assignment 2 data structures description: • The RAM o Implemented as an array of FILE pointers. ▪ FILE *ram[10]; o Each program executed by the shell’s exec command is loaded into ram[]. This is not true for the mysh scripts, they are loaded and executed as in assignment 1. The exec command runs programs concurrently, therefore the programs need to be in ram[] at the same time. To simplify this, RAM is an array of FILE pointers. This RAM has space for 10 FILE pointers, which relates to 10 programs. A program is said to be in memory when its file pointer occupies one of the cells of ram[]. A NULL pointer indicates that there is no program in ram[]. o When a program is loaded to ram[] the following operations happen: Operating Systems COMP 310 – ECSE 427 McGill University Vybihal Assignment #2 Page 3 of 6 ▪ Step 1: fopen the script ▪ Step 2: find the next available cell in ram[] ▪ Step 3: assign the FILE pointer from the fopen to that cell ▪ Notes: • To read a line from the program do: fgets(ram[k],buffer,limit); Where k is the current executing program. • When the program has finished executing do : fclose(ram[k])) and then ram[k]=NULL; • The Ready Queue o Implemented as a simple linked list with head and tail pointers. ▪ PCB *head, *tail; o The ready queue is FIFO and RR. Pointer “head” points to the first PCB in the list. The first PCB is the one that gets the CPU. Pointer “tail” points to the last PCB in the list. New PCBs are appended at the tail. Make sure to update “tail” to point to the new PCB after appending it to the list. • The PCB o For this assignment our PCB will be very simple. It will contain only a program counter. Since the executed programs are files, the program counter will be a file pointer. ▪ struct PCB { FILE *PC; }; o Note: the PCB PC filed is not the CPU instruction pointer, so only at each task switch does the PCB PC field get updated. o Note: when a program is launched, a PCB is created, and the PCB PC field points to the first line of the program. This means at program launch the PCB PC field and the ram[] FILE pointer are pointing to the beginning of the file. This will change after the next quanta. The ram[] FILE pointer will still point to the beginning of the file, but the PCB PC field will point to the current line to be executed the next time the PCB gets the CPU. • The CPU o The CPU is simulated by a struct having an instruction pointer (IP), instruction register (IR), and a quanta field. The IP is like the PCB PC field, it points to the next instruction to execute from the file. The currently executing instruction is stored in the IR. ▪ struct CPU { FILE *IP; char IR[1000]; int quanta=2; } o Note: the IR stores the instruction that will be sent to the Interpreter() for execution. o Note: since we are not implementing threads, all concurrent programs access the same shell memory space. o Note: for this assignment we assume the quanta is 2 lines of code for each program. Assignment 2 operation of the exec command: The user, at the command line, can input any of the following: exec prog.txt exec prog.txt prog2.txt exec prog.txt prog2.txt prog3.txt Operating Systems COMP 310 – ECSE 427 McGill University Vybihal Assignment #2 Page 4 of 6 In the first case only one program is launched in the kernel. In the second case two programs are launched. In the third case three programs are launched. Launched means the following: (1) each file is fopened and the FILE pointer is assigned to a cell in ram[]. (2) a PCB is created (malloc) for that program and the same FILE pointer that was assigned to ram[] is assigned to the PCB PC field. (3) the PCB is added to the Ready Queue. This is done for all the programs in the list. Since we are not implementing threads the shell prompt only displays after the exec command has finished executing all the programs. This is how a program completes execution: (a) the ram[] cell is assigned NULL, (b) the PCB is removed from the Ready Queue, and (c) free(PCB) to release C language memory. Assignment 2 program execution: Only after the “exec” command loads all the programs into ram[] and appends the PCBs to the Ready Queue, do the programs start to run. Each PCB in the Ready Queue is sorted First-in First- out (FIFO) and Round Robin (RR), in the same order as they appeared in the exec command. The exec() function is an interpreter function stored in interpreter.c. It handles the filename argument verification error. It opens each program file. It calls the kernel function to load each program into the simulation. The final thing it does, it starts the execution of all the loaded
Answered Same DayFeb 23, 2021

Answer To: Operating Systems COMP 310 – ECSE 427 McGill University Vybihal Assignment #2 Page 1 of 6 Assignment...

Pratik answered on Feb 25 2021
132 Votes
Kernel/kernel.h
#include
//#include "pcb.h"
void addToReady(struct PCB *p);
void myinit(FILE *p);
void scheduler();
void initKernel();
void init();
Kernel/mykernel
Kernel/README.md
#KERNEL
The folder contains all the files required for a kernel.
To build the kernel use the command -
gcc -o mykernel
shell.c interpreter.c shellmemory.c kernel.c cpu.c pcb.c ram.c
##Run the program
Usage - ./mykernel
The folder also contains TESTFILE.txt, it can be used as
./mykernel < TESTFILE.txt
##Learning Programming
Please refer to the main() function in the kernel.c file
Kernel/prog2.txt
set a 100
print b
set b 250
print b
Kernel/ram.h
#include
void initRAM();
void addToRAM(FILE *p);
void removeMemory(FILE *p);
Kernel/interpreter.c
#include
#include
#include
#include "shellmemory.h"
#include "shell.h"
#include "kernel.h"
int run(char *filename) {
    FILE *ptr;
    char buffer[1000], buf0[100], buf1[100], buf2[100];
    int result = 0;
    ptr = fopen(filename,"rt");
    if (ptr == NULL) return 4; // file not found
    fgets(buffer,999,ptr);
    while(!feof(ptr)) {
        if (strlen(buffer)>1) result = prompt(buffer);
        if (result == 99) break;
        fgets(buffer,999,ptr);
    }
    fclose(ptr);
    if (result == 99) return 99;
    return 0;
}
int exec(char buf1[],char buf2[],char buf3[])
{
    FILE *ptr1,*ptr2,*ptr3;
    ptr1=ptr2=ptr3=NULL;
    if(strcmp(buf1,buf2)==0)
    {
        printf("Error : Script %s already loaded\n",buf1);
        buf2="\0";
    }
    else if(strcmp(buf1,buf3)==0)
    {
        printf("Error : Script %s already loaded\n",buf1);
        buf3="\0";
    }
    else if((strlen(buf2)>0)&&(strcmp(buf2,buf3)==0))
    {
        printf("Error : Script %s already loaded\n",buf2);
        buf3="\0";
    }
    ptr1 = fopen(buf1,"rt");
    if(strlen(buf2)>1)
        ptr2 = fopen(buf2,"rt");
    if(strlen(buf3)>1)
        ptr3 = fopen(buf3,"rt");
    if(ptr1==NULL) return 4;
    myinit(ptr1);
    if(ptr2!=NULL)
    myinit(ptr2);
    if(ptr3!=NULL)
    myinit(ptr3);
    scheduler();
    return 0;
}
int interpreter(char buf0[], char buf1[], char buf2[],char buf3[]) {
    int result = 0; // no errors
    if (strcmp(buf0,"help")==0) {
        printf("Legal commands:\n");
        printf("help display this help\n");
        printf("quit exits the shell\n");
        printf("set VAR STRING assign STRING to VAR\n");
        printf("print VAR display contents of VAR\n");
        printf("run SCRIPT.TXT interpret SCRIPT.TXT\n");
        printf("exec prog1.txt [prog2.txt] [prog3.txt]\n");
        result = 0;
    }
    else if (strcmp(buf0,"quit")==0) {
        result = 99; // exit shell code
    }
    else if (strcmp(buf0,"set")==0) {
        if (strlen(buf1)<1 || strlen(buf2)<1) return 1; // set error
        add(strdup(buf1), strdup(buf2));
    }
    else if (strcmp(buf0,"print")==0) {
        if (strlen(buf1)<1) return 2; // print error
        printf("%s\n", get(buf1));
    }
    else if (strcmp(buf0,"run")==0) {
        if (strlen(buf1)<1) return 3; // run error
        result = run(buf1);
    }
    else if(strcmp(buf0,"exec")==0){
        if (strlen(buf1)<1) return 5; // exec error
        
        result = exec(buf1,buf2,buf3);
    }
    else {
        result = 98; // command does not exist
    }
    return result;
}
Kernel/shell.c
#include
#include
#include...
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