cs609 Assignment No.2 solution&discussion Due date: 2 December 2019

[ayesha]

	Semester: Fall 2019 CS609: System Programming Graded Assignment No. 02		Total Marks: 20   Due Date:Dec02, 2019
Instructions:   Please read the following instructions carefully before submitting assignment. It should be clear that your assignment will not get any credit if:   §  The assignment is submitted after due date. §  The submitted assignment does not open or file is corrupt. §  You have not followed steps described in Detailed Instructions of the problem statement. §  Assignment is copied(partial or full) from any source (websites, forums, students, etc.) Strict action will be taken in this regard.   Note:You have to upload only .doc or .docx file. Assignment in any other format (extension) will not be accepted and will be awarded with zero marks.   Objectives:   The objective of this assignment is to provide hands-on experience ofSystem Programming concepts including:   ·         How can interrupts be generated ·         What are Interrupts ·         Interrupt functions writing ·         TSR program ·         Calling interrupt functions             For any assignment related query, contact at [e-mail]
Problem Statement: You are required to write C programwhich will display one ‘*’ characteron the screen on each CPU timer interrupt 8 in a way that each * character is displayed after the previous *. i.e ********************************************  On elapse of 7 seconds then the screen characters should be removed and again the * character should be started printing on the screen in similar way i.e ********* This process should be continued again and again until you press any key from the keyboard. Instructions: 1)     You should include all related header files first of all then declare interrupt pointer to hold Timer i.e void interrupt (*oldTimer)(void); 2)     Similarly, give prototype for new functions of newtimer()i.e void interrupt newTimer(); 3)     Declare a far pointer i.e *scr to hold far address =0xB8000000; 4)     Store current vector values of INT 8 through getvect in oldTimer. 5)     Set newTimer fuctions through setvect. 6)     In newTimer() function, give logic to print ‘*’ on the screen i.e *(scr+i)=0x2A; 7)     Write code to wait for 7 seconds i.e t>=126and through loop write blank spaces on the screen by  setting keyboard status with black background i.e no character on screen i.e *(scr+i)=ox20; *(scr+i+1)=ox07; 8)     When any key is pressed from the keyboard, the program should stop. Note:Your assignment should Word file (.doc or .docx) containing your code.   Best of Luck!

 

[+ M.Tariq Malik]

CS609_Assignmnet_2_Solution

below link 

CS609_Assignmnet_2_Solution.txt

[MIAN MUHAMMAD USMAN]

Tariq bhai sahi solution upload kro please

[MIAN MUHAMMAD USMAN]

Please right solution upload kro. 

[Muhammad Asif]

void interrupt (*oldTimer)(void);
void interrupt (*oldKey)(void);
void interrupt newTimer();
void interrupt newKey();
char far *scr = (char *far) 0xB8000000;
int i, t=0, m=0;
void main()
{
clrscr();
oldTimer = getvect(8);
oldKey =getvect(9);
setvect(8,newTimer);
setvect(9,newKey);
getch();
}

void interrupt newTimer()
{
t++;
if (t<=126) { for(i=0;i<=t;i+=2) { *(scr+i)=0x2A; } } if(t>=126)
{
for(i=0;i<=4000;i+=2) { *(scr+i)=0x20; *(scr+i+1)=0x07; } t=0; m=1; } (*oldTimer)(); } void interrupt newKey() { if (m==1) { m=0; exit(0); } (*oldKey)(); }

[+ M.Tariq Malik]

CS609 Assignment No 02 Solution:

#include<stdio.h>
#include<stdlib.h>
#include<DOS.H>
#include<BIOS.H>
#include<conio.h>
void interrupt (*oldTimer)(void);
void interrupt (*oldKey(void);
void interuupt newTimer();
void interrupt (newKey(void);
char far *scr = (char *far) 0xB8000000;
int i, t=0, m=0;
char charscr[4000]={"*"};
void main()
{
oldTimer = getvect(8);
oldKey =getvect(9);
setvect(8,newTimer);
setvect(9,newKey);
getch();
getch();
getch();
getch();
}
void interrupt newTimer()
{
t++;
if ((t>=126) && (m==0))
{
for (i=0;i<4000;i++)
*(scr+i)=0x2A;
for(i=0;i<=4000;i+=2)
{
*(scr+i)=0x20;
*(scr+i+1)=0x07;
}
t=0;
m=1;
}
(*oldTimer)();
}
void interrupt newLKey()
{
if (m==1)
{
m=0;
exit(0);
}
(*oldKey)();
}