Program for enabling and disabling the keyboard.

#include "stdio.h"
#include "conio.h"
#include "dos.h"

main()
{
clrscr();
outportb(0x64,0xAD); /*Code for disabling the KEYBOARD*/
printf("Keyboard Disabled");
delay(10000);
clrscr();
printf("Keyboard Enabled");
outportb(0x64,0xAE); /*Code for enabling the KEYBOARD*/
getch();

}

Program for verifying the disk sector.

#include "stdio.h"
#include "conio.h"
#include "dos.h"

void main(void)
{
clrscr();
union REGS regs;
struct SREGS sregs;
char buff[1000];
int i;
regs.h.ah = 4;
regs.h.al = 1;
regs.h.ch = 1;
regs.h.dh = 0;
regs.h.cl = 1;
regs.h.dl = 0x80;
regs.x.bx = FP_OFF(buff);
sregs.es = FP_SEG(buff);
int86x(0x13,®s,®s,&sregs);
printf("regs.x.cflag - %d",regs.x.cflag);
printf("regs.h.ah - %d",regs.h.ah);
printf("regs.h.al - %d",regs.h.al);
printf("Buffer - %d",buff);
getch();
}

Program To transfer text and html files between two computers using three wire connections between com ports

#include"stdio.h"
#include"conio.h"
#include"process.h"
#include"dos.h"
#include"graphics.h"
#include"start.h"

void baudset(long);
void comparm(int,int,int);
int port_address;
union REGS i,o;
void main()
{
int gd=DETECT,gm,maxx,maxy,x,y,button,ex=1,c,v=1,run=1;
FILE *f;
char chr;
char *fname;
int result,status,choice,port;
unsigned int fstat,mstat,lstat;
st();
while(1)
{
initgraph(&gd,&gm," ");
cleardevice();
maxx=getmaxx();
maxy=getmaxy();
setbkcolor(9);
if(run==1)
{
gotoxy(32,1);
printf("PORT INACTIVE");
run++;
}
else
{
gotoxy(32,1);
printf("PORT ACTIVATED SUCCESSFULLY");
}

choice=0;
rectangle(0,20,maxx,maxy);
setviewport(1,5,maxx-1,maxy-1,1);
if(initmouse()==0)
{
closegraph();
restorecrtmode();
outtextxy(100,100,"mouse driver not loaded");
exit(1);
}
restrictmouseptr(1,20,maxx-1,maxy-1);
showmouseptr();
setcolor(6);
outtextxy(115,30,"WELCOME TO PROGRAM TO TRANSFER FILES BETWEEN TWO COMPUTERS");
setcolor(3);
outtextxy(430,45,"ver 1.0 ");
setcolor(14);
outtextxy(170,45,"PANKAJ A.H., 9373144662");
setcolor(10);
outtextxy(200,100,"1 : SEND FILE");
outtextxy(200,120,"2 : RECIEVE FILE");
outtextxy(200,140,"3 : EXIT");
outtextxy(200,175,"enter your choice (1,2 or 3)::");
rectangle(195,95,210,110);
rectangle(195,115,210,130);
rectangle(195,135,210,150);
setcolor(13);
outtextxy(20,450,"PRESS 'F1' TO ACTIVATE THE PORTS");
setcolor(9);
while(v==1)
{
if(kbhit())
{
chr=getch();
if(chr==0)
{
chr=getch();
c=(int)chr;
if(c==59)
{
gotoxy(31,1);
printf("PORT ACTIVATED");
v=2;
break;

}
}
}
}
while(ex==1)
{
getmousepos(&button,&x,&y);
gotoxy(5,1);
(button & 1)== 1 ? printf("DOWN"):printf("UP");

gotoxy(20,1);
(button & 2) == 2 ? printf("DOWN"):printf("UP");

gotoxy(65,1);
printf("X=%03d Y=%03d",x,y);
if((button & 1)==1)
{
if((x>=195 && x<=210) && (y>=95 && y<=110))
{
choice=1;
}
if((x>=195 && x<=210) && (y>=115 && y<= 130))
{
choice=2;
}
if((x>=195 && x<=210) && (y>=135 && y<=150))
{
choice=3;
}
}
if(choice==1 || choice==2 || choice==3)
goto sos;
if(kbhit())
{
goto sob;
}
}
fflush(stdin);
sob: gotoxy(60,12);
scanf("%d",&choice);
sos: switch(choice)
{
case 1: {
setcolor(12);
outtextxy(10,205,"select port (1: com1 2: com2) ::");
rectangle(470,200,485,215);
rectangle(510,200,525,215);
outtextxy(474,203,"1");
outtextxy(514,203,"2");
gotoxy(50,14);
scanf("%d",&port);
if(port==1)
port_address=0x3F8;
else
port_address=0x2F8;
baudset((long)2400);
comparm(0,1,8);
outport(port_address+4,1|2);
outtextxy(10,240,"enter the name of file to be sent::");
gotoxy(50,16);
scanf("%15s",&fname);
f=fopen(&fname,"r");
if(f==NULL)
{
setcolor(4);
outtextxy(10,300,"*** file do not exist ***");
setcolor(6);
outtextxy(10,320,"hit any key to continue........");
getch();
break;
}
closegraph();
restorecrtmode();
clrscr();
textcolor(2);
for(;;)
{
if(kbhit())
{
if(getch()==27)
break;
}
lstat=inportb(port_address+5);
if(!(lstat & 32))
continue;
else
{
chr=fgetc(f);
}
if(chr==EOF)
{
cprintf("FILE TRANSMITTED
);
outportb(port_address,EOF);
getch();
break;
}
outportb(port_address,chr);
putchar(chr);
}

fclose(f);
break;
}

case 2: {
setcolor(12);
outtextxy(10,205,"select port (1: com1 2: com2) ::");
gotoxy(50,14);
scanf("%d",&port);
if(port==1)
port_address=0x3F8;
else
port_address=0x2F8;
baudset((long)2400);
comparm(0,1,8);
outportb(port_address+4,1|2);
textcolor(4);
outtextxy(10,240,"save as file name::");
gotoxy(50,16);
scanf("%15s",&fname);

f=fopen(&fname,"w");
if(f==NULL)
{
setcolor(4);
outtextxy(10,300,"*** sorry cannot open file ***");
break;
}
closegraph();
restorecrtmode();
clrscr();
textcolor(4);
cprintf("waiting for i/p
);
textcolor(7);
cprintf("press esc for exit
);
gotoxy(1,4);
for(;;)
{
if(kbhit())
{
if(getch()==27)
break;
}
fstat=inportb(port_address+5);
if(fstat & 2)
cprintf("over run time error
);
if(fstat & 4)
cprintf("parity error");
if(fstat & 8)
cprintf("framing error
);
if(fstat & 16)
cprintf("break recieved
);
if(!(fstat & 1))
continue;
chr=inportb(port_address);
if(chr==EOF)
{
textcolor(2);
cprintf("FILE RECIEVED
);
exit(0);
}
putchar(chr);
fputc(chr,f);
}
outportb(port_address+4,0);
fclose(f);
break;
}

case 3: exit(0);
default: sound(900);
delay(10000);
nosound();
break;
}

}

}


void baudset(long baudrate)
{
unsigned int divisor;
unsigned char lsb,msb;
divisor=1152001/baudrate;
msb=divisor>>8;
lsb=(divisor<<8)>>8;
outportb(port_address+3,128);
outportb(port_address,lsb);
outportb(port_address+1,msb);

}


void comparm(int parity,int stop,int databit)
{
int parmbyte;
parmbyte=databit-5;
if(stop==2)
parmbyte|=4;
if(parity!=0)
parmbyte|=8;
if(parity==2)
parmbyte|=16;
outportb(port_address+3,parmbyte);
}

/* MOUSE WORK */

initmouse()
{
i.x.ax=0;
int86(0x33,&i,&o);
return(o.x.ax);
}


showmouseptr()
{
i.x.ax=1;
int86(0x33,&i,&o);
}


restrictmouseptr(int x1,int y1,int x2,int y2)
{
i.x.ax=7;
i.x.cx=x1;
i.x.dx=x2;
int86(0x33,&i,&o);
i.x.ax=8;
i.x.cx=y1;
i.x.dx=y2;
int86(0x33,&i,&o);
}


getmousepos(int *button,int *x,int *y)
{
i.x.ax=3;
int86(0x33,&i,&o);
*button=o.x.bx;
*x=o.x.cx;
*y=o.x.dx;
}

Expression of a human face which smiles and cries

#include "graphics.h"
#include "stdlib.h"
#include "stdio.h"
#include "malloc.h"
#include "dos.h"
#include "conio.h"

int xasp,yasp,gdriver = VGA, gmode=VGAMED, errorcode;

struct pos
{
int x;
int y;
};

struct face
{
int radius;
struct pos position;
int mood;
};

typedef struct face face;
face *face1;

void getposition()
{
printf("Enter X Co-ordinate:");
scanf("%d",&face1->position.x);
printf("Enter X Co-ordinate:");
scanf("%d",&face1->position.y);
}

void drawface()
{

char ch='x';
int i=0,x,y,color,r,imgsize,dif;

x=face1->position.x=320;
y=face1->position.y=180;
face1->radius=150;
color=15;
r=face1->radius;

setbkcolor(0);
getaspectratio(&xasp,&yasp);
setcolor(8);
circle(x,y,face1->radius);

setfillstyle(1,color);
floodfill(x,y,getcolor());


draweyes(face1);
drawhair(face1);
drawmouth(face1);
drawnose(face1);


}

drawnose()
{
int i,x,y,r;
x=face1->position.x;
y=face1->position.y;
r=face1->radius;
setcolor(0);

for(i=0;i<2;i++) r="face1-">radius;
x1=face1->position.x-r/2;
y1=face1->position.y-r/4;
x2=face1->position.x+r/2;
y2=face1->position.y-r/4;

setaspectratio(xasp/2,yasp);
arc(x1,y1-r/8,40,140,r/4);//left eyebrow
arc(x1,y1-r/8+1,40,140,r/4);//left eyebrow
arc(x1,y1-r/8+2,40,140,r/4);//left eyebrow
setaspectratio(xasp,yasp);

for(i=0;i<2;i++) i="0;i<2;i++)" x="face1-">position.x;
y=face1->position.y+(face1->radius/1.5);
r=face1->radius;
setcolor(BLACK);

if((face1->mood)==1)
for(i=0;i<4;i++)>mood)==0)
for(i=0;i<4;i++) r="face1-">radius;
x=face1->position.x-r/2;
y=face1->position.y-r/3;

arc(x,y,34,225,100);

arc(x+r,y,314,138,100);

setfillstyle(1,RED);
floodfill(x,y-70,getcolor());
floodfill(x+r,y-70,getcolor());
setaspectratio(xasp,yasp);
}

void main(void)
{
int i=0;
initgraph(&gdriver, &gmode,"");

while(!kbhit())
{
if((i%2)==1)
{
setvisualpage(1);
setactivepage(0);
clearviewport();
face1->mood=0;
drawface();
delay(1000);
}
else
{
setvisualpage(0);
setactivepage(1);
clearviewport();
face1->mood=1;
drawface();
delay(300);
}
i++;
}

getch();
closegraph();
}

Convert between bases

#include
#include
#include
#include

int main()
{
int cotinue=1;
long int org_num=0, org_base=0, valid_org_val, pos_org_num, new_num, base10_num, remainder;
int new_base=0;
int pow_limit, pow_test;
int cur_term=0;
cout<
do
{
//The outermost "do...while" loop allows for repetitive revolutions of
//the primary block of the application. The program will continue to
//oscillate between the input and output pages as long as the variable
//referred to as "cotinue" is equal to "1." This variable can be changed
//through user input at the end of the output page.
do
{
//This repetition statement ensures that the base of the initial number,
//which is discovered by the program through a user input statement,
//is a value between 1 and 17 exclusively. The prompt message will
//continue to be displayed and the user will continue to be required
//to enter in the value of the base for the original number if he/she
//incessently enters invalid bases into the system.
cout<
cin>>new_base;
} while((new_base <> 16));
do
{
//This "do...while" loop will continue to run as long as the "org_num"
//variable, which sybolizes the amount of the initial number, does not
//meet specific requirements concerning the size of the first value. The
//user is expected to abide by the rules connecting the value of the
//new base and the amount of the number that must be converted,
//which are explained to the operator during the execution of the project
//with the aid of the "cout" statement below. If the first amount is too
//large or too small (in terms of a negative value) to be converted
//into the specified base, the user will be required to enter in a new
//number which fits the parameters for the new base.
cout<<"\n\nInput the integer that must be converted.\n\nIf the new base is 2, then the original number must be\ngreater than or equal to -1,000 and less than or equal to 1,000.\n\nIf the new base is > 2 and <>
cin>>org_num;
} while((((org_num < -1000) || (org_num > 1000)) && (new_base == 2)) || (((org_num < -10000) || (org_num > 10000)) && ((new_base > 2) && (new_base <> 1000000)) && (new_base > 10)));
cout<
do
{
//The nested "do...while" repetition statement will repeat as long as
//the original base of the original integer is less than 2 or greater
//than 10. Since the variable of "org_num," which stores the initial
//whole number, is an integer, the base of the first number cannot exceed
//10. If the base was allowed to exceed 10, then letters and other
//nonnumerical characters would be allowed as input into the "org_num"
//integer identifier, effectively breaking C++ syntax rules. Similar
//to the two previous "do...while" loops, this repetition unit will
//continue to execute with a prompt for user input continually being
//displayed on the monitor while the base of the starting whole number
//is less than 2 or larger than 10.
cout<<"\nWhat is the base of the initial number\n"<
cin>>org_base;
} while((org_base <> 10));
clrscr(); //Second Page Begins to Be Processed
//***Full Conversion Process Begins Below
if(org_num <>
//If it is discovered that the original number is negative, the initial amount
//is transformed into a positive value by storing the product of -1 and
//"org_num" in the identifier named "pos_org_num."
pos_org_num = org_num * -1;
else
//If "org_num" is positive or equal to 0, the value of the first whole number
//is placed in "pos_org_num."
pos_org_num = org_num;
//**Start of Conversion to Base Ten
for(pow_test=0; pow(10, pow_test) <= pos_org_num; ++pow_test);
//The above "for" loop, which does not contain a single executable statement
//but rather serves as a counter, determines the final value of "pow_test."
//"pow_test" is an integer variable that holds the first value "n" in
//the expression "1 * 10^n" such that the result of the expression
//is greater than "pos_org_num." The final value of "pow_test" also
//stores the number of single digits that are found in the positive
//form of the intial number. "pow_test" is incremented by 1 after each
//revolution through the loop, and the repetition statement will end
//when the result of "10^pow_test" is greater than the original amount
//in its positive form.
for(pow_limit=(pow_test - 1), remainder=pos_org_num, base10_num=0, valid_org_val=1; pow_limit >= 0; --pow_limit)
{
//The purpose of this "for" loop involves converting the initial amount,
//which is now present in its positive form in the variable called
//"pos_org_num," to its equivalent value in base 10. The identifier
//named "pow_limit" is initialized at an upper bound--the number of
//digits that are in "pos_org_num" subtracted by 1. "remainder,"
//an identifier that stores the amount within "pos_org_num" which have
//not been subtracted by the succeeding processes, is set to the positive
//form of "org_num." "base10_num" is an accumulator that holds the current
//result of the conversion process of "pos_org_num" into a base 10 value.
//On the other hand, "valid_org_val" is initialized to 1, signifying
//that the program assumes at first that all digits in "pos_org_num"
//are less than the original base that the value is in. This loop will
//continue to execute as long as "pow_limit" is greater that -1, and
//the value stored in "pow_limit" will decrement by 1 at the end of each
//revolution.
cur_term = (remainder / pow(10, pow_limit));
//The above assignment statement allows the computer to identify any
//digit that lies within "pos_org_num."
if(cur_term >= org_base)
{
//If the current term that is being analyzed in the "for" loop is
//discovered to be greater than or equal to the base that the amount
//was originally in, "pow_limit" is set to -1 (the outer "for" loop
//will terminate when "pow_limit" is less than 0) and "valid_org_base"
//is assigned the value of 0, signifying that the orginal value does
//not coincide with its suggested original base.
pow_limit=-1;
valid_org_val=0;
}
base10_num = (base10_num + (cur_term * pow(org_base, pow_limit)));
//The previous value of "base10_num" is added to the current digit that
//was pulled from the positive form of the initial number times the result
//of the number's original base raised to the current value of "pow_limit."
//After the surrounding repetition statement is finished executing,
//"base10_num" will contain the value of the original amount converted
//into base 10 number representation. This process is performed to
//improve the logical pathway involved in transforming a number in any
//base that is less than 10 into its equivalent amount in a second base
//that is under 10.
remainder = (remainder - (pow(10, pow_limit) * cur_term));
//The updated version of "remainder" is found by raising 10 to the current
//value for "pow_limit," multiplying the result by the digit that is being
//analyzed from the intial amount, and subtracting the product from the
//previous value of "remainder."
}
//**End of Base Ten Conversion
if(valid_org_val == 0)
//If the original number contains digits that exceed or equal the first
//number's base, an error message will be outputted and the program will
//ask the operator if he/she wants to attempt the conversion stage again.
cout<<"\n\n\n\nError: The Base and the Value of the Original Number Do not Coincide";
else
{
//This block of code will be accessed by the computer if all of the digits
//in the initial number are less than the base that the number is supposedly in.
cout<<"\n\n\n\nThe conversion of "<
for(pow_test=0; pow(new_base, pow_test) <= base10_num; ++pow_test);
//Although the above "for" loop lacks a body, its heading serves as
//a counter for the number of characters that lie within the newly
//created "base10_num" variable. The loop's control variable, which
//is referred to as "pow_test," will increment by one after each run
//through the loop, and the loop will continue to execute while the
//result of "new_base ^ pow_test" is less than or equal to the base 10
//conversion of the positive form of the original number.
if(new_base <>
{
//If the base that the intial value is being converted over to is
//less than 11, then the following segment of code is performed.
for(pow_limit=(pow_test-1), remainder=base10_num, new_num=0; pow_limit >= 0; --pow_limit)
{
//The purpose of this repetition statement is to convert the newly
//developed "base10_num" variable into the new base that was
//specified by the user.
//This "for" loop initializes "pow_limit," which symbolizes the digit
//that is being reviewed within "base10_num" in terms of scientific
//notation, to the difference of "pow_test" minus 1. In addition,
//"remainder," a variable that holds the amount in "base10_num"
//that has not been subtracted from the variable's original value
//by the following operations, is first set to equal "base10_num."
//The sum of the conversion process of "base10_num" to the number's
//new base is set to a starting value of 0. The loop's control variable
//of "pow_limit" decreases by 1 following each successful cycle of the
//repetition statement's body, and the loop's body will continue
//to execute as long as "pow_limit" is not less than 0.
cur_term = pow(new_base, pow_limit);
//The current term or digit that will be divided by "base10_num"
//is calculated and stored in "cur_term" by raising the original
//integer's new base to the value of "pow_limit."
new_num = new_num + (pow(10, pow_limit) * (remainder / cur_term));
//The value of "new_num," which represents the current progress of
//converting the original whole number to the new base, is found
//by raising 10 to "pow_limit," multiplying the result by the outcome
//of "remainder" divided by "cur_term," and adding the product to
//the previous value of "new_num."
remainder = (remainder - ((remainder / cur_term) * cur_term));
//The new value of "remainder" stores the current amount residing
//in "base10_num" which have not been "touched" by the division
//and conversion processes.
}
if(org_num <>
//If the original amount is discovered to be negative, the positive
//form of "new_num" will become negative by multiplying "new_num"
//by -1.
new_num = new_num * -1;
cout<
}
else
{
//If the new base is not between 2 and 10, inclusively, then an entirely
//different process must be performed to reveal the proper answer.
//Instead of storing the converted solution to the original number in
//a variable, the individual characters, including letters and numbers,
//are printed on the monitor as they are discovered.
for(pow_limit=(pow_test-1), remainder=base10_num; pow_limit >= 0; --pow_limit)
{
//This "for" loop is designed to calculate and display each value
//of the conversion's solution character-by-character. The loop's
//control variable (or "pow_limit") is initialized at "pow_test - 1,"
//which reveals the number of digits that exist in the number stored
//in "base10_num." On the other hand, "remainder" is first set to
//equal "base10_num." The loop's control variable decrements by 1
//at the conclusion of each revolution through the loop, and the
//loop will run as long as "pow_limit" is not less than 0.
cur_term = pow(new_base, pow_limit);
//The current "term" of the conversion process is calculated by
//raising the new base to the current value for "pow_limit." For
//instance, if "new_base" is 6 and "pow_limit" is equivalent to
//3, then "cur_term" equals 216.
if((remainder / cur_term) <>
//If the newly discovered character in the solution is a number
//that is less than 10, then the numerical form of the single
//digit is displayed on the screen with the aid of the
//output statement that is written below.
cout<<(remainder / cur_term);
else
//If the new character in the problem's solution is a value
//which is greater than 9, then the letter form of the new number
//is displayed to the user. For example, if the new element is
//12, the letter "C" will be displayed instead of "12."
cout<
remainder = (remainder - ((remainder / cur_term) * cur_term));
//The above line of code determines the new value of "remainder,"
//which contains the amount of "base10_num" that has not been affected
//by the program.
}
}
}
//***End of Full Base Conversion
cout<<"\n\n\n\n\n\n\n\tWould you like to run the conversion process again?\n\t\t\t\t(1 = Yes, 2 = No)\n\t\t\t\t\t";
//The above output statement and the input line below require the user to
//choose whether he/she would like to cycle through the base conversion
//process again. If the operator enters a "1," the first screen of the
//application will reappear. If the user presses any key other than "1,"
//the project will terminate.
cin>>cotinue;
clrscr();
} while(cotinue == 1);
return 0;
}

Showing Binary Equivalent of an Integer

main()
{
int n;
char ch;
while(1)
{
clrscr();
printf("Enter an integer to see its Binary Equivalent");
printf("Enter ");
scanf("%d",&n);
showbits(n);

printf("continue ? ");
ch=getch();
if(ch==27)
break;
}
}

showbits(int n)
{
int i,bit,mask,count=0;
printf(" ");
for(i=15;i>=0;i--)
{
mask=1<

bit=n & mask;

if(bit==mask)
printf("1");
else
printf("0");
count++;
if(count%4==0)
printf(" ");
}
}

A Program to find permutation in c

#include "stdio.h"
#include "stdlib.h"
int lev=-1,n,val[50],a[50];
void main()
{
int i,j;
clrscr();
printf("Enter howmany numbers ");
scanf("%d",&n);
for(i=0;i
{
val[i]=0;
j=i+1;
scanf("%d",&a[j]);
}
visit(0);
getch();
}
visit(int k)
{
int i;
val[k]=++lev;
if(lev==n)
{
for(i=0;i
printf("%2d",a[val[i]]);
printf(" ");
}
for(i=0;i
if(val[i]==0)
visit(i);
lev--;
val[k]=0;
}

Program to Create n-sized triangle.

/* Print a Triangle of Stars on it's tip
Input: a positive integer
Output: an n-sized triangle */
void Triangle(int s)
{
int i = 0;

while (s > 0)
{
for (i = 0; i < s; i++)
{
printf("*");
}
printf("\n");
s = s - 1;
}
}

/* Print a Triangle of Stars on it's base
Input: a positive integer
Output: an n-sized triangle */
void TriangleTwo(int t)
{
int i = 0;
int x = 0;
int y = 1;
int counter = 0;

while (counter < t)
{
x = t - (t - y);

for (i = 0; i < x; i++)
{
printf("*");
}
printf("\n");
y++;
counter++;
}
}




**************


void TriangleTwo(int t)
{
if (t < 1)
return; /* base case */
else
{
TriangleTwo(t - 1); /* recursive step */

while(t > 0)
{
printf("*");
t--;
}
printf("\n");

}
}





*****************

void printIt(int row, int star)
{
if(row)
{
putchar('*');
--star;
if(star == 0)
{
putchar('\n');
star = row - 1;
--row;
}
printIt(row, star);
--row;
}
else
{
return;
}
}



**********************

// Print n asterisks and a newline
void asterisks(int n)
{
if (n <= 0)
putchar('\n');
else
{
putchar('*');
asterisks(n - 1);
}
}

// Descending triangle (large at top, small at bottom)
void t1(int n)
{
if (n > 0)
{
asterisks(n);
t1(n - 1);
}
}

// Ascending triangle (small at top, large at bottom)
void t2(int n)
{
if (n > 0)
{
t2(n - 1);
asterisks(n);
}
}


**********************



/* Print a Triangle of Stars on it's tip
Input: a positive integer
Output: an n-sized triangle */
void Triangle(int s)
{

if (s == 1)
printf("*"); /* base case */
else
{
for(int i = 1; i <= s; i++)
{
printf("*");
}
printf("\n");
Triangle(s - 1); /* recursive step */
}
}

/* Print a Triangle of Stars on it's base
Input: a positive integer
Output: an n-sized triangle */
void TriangleTwo(int t)
{
if (t < 1)
return; /* base case */
else
{
TriangleTwo(t - 1); /* recursive step */

while(t > 0)
{
printf("*");
t--;
}
printf("\n");

}
}

Employee Project in C

#include "stdio.h"
main()
{
FILE*fp,*ft;
char another,choice;

struct emp
{
char name [20];
int age;
float bs;
};

struct emp e;
char empname[40];
long int recsize;
fp=fopen("EMP.DAT","rb+");

if(fp==NULL)
{
fp=fopen("EMP.DAT","wb+");
exit();
}

recsize=sizeof(e);

while(1)
{
clrscr();
printf("\n 1.Add records");
printf("\n 2.List records");
printf("\n 3.Modifay records");
printf("\n 4.Delete records");
printf("\n 0.Exit");
printf("\n Enter your choice");

fflush (stdin);
choice=getche();

switch(choice)
{
case '1':
fseek(fp,0,SEEK_END);
another='y';
while(another=='y')
{
printf("\n enter name age and basic salary");
scanf("%s %d %f",&e.name,&e.age,&e.bs);
fwrite(&e,recsize,1,fp);
printf("\n add another records(y/n)");
fflush(stdin);
another=getche();
}
break;

case '2':
rewind(fp);
while(fread(&e,recsize,1,fp)==1)
printf("\n %s %d %f",e.name,e.age,e.bs);
getch();
break;

case '3':
another='y';
while(another=='y')
{
printf("\n enter name of employee to modify");
scanf("%s",&empname);

rewind(fp);
while(fread(&e,recsize,1,fp)==1)
{
if(strcmp(e.name,empname)==0)
{
printf("\n enter new name age and bs");
scanf("%s %d %f",&e.name,&e.age,&e.bs);
fseek(fp,-recsize,SEEK_CUR);
fwrite(&e,recsize,1,fp);
break;
}
}
printf("\nmodify another record(y/n)");
fflush(stdin);
another=getche();
}
break;

case '4':

another ='y';
while(another=='y')
{
printf("\n enter name of employee to delete");
scanf("%s",&empname);
ft=fopen("TEMP.DAT","wb");
rewind(fp);
while(fread(&e,recsize,1,fp)==1)
{
if(strcmp(e.name,empname)!=0)
fwrite(&e,recsize,1,ft);
}
fclose(fp);
fclose(ft);

remove("EMP.DAT");
rename("TEMP.DAT","EMP.DAT");

fp=fopen("EMP.DAT","rb+");
printf("delete another record(y/n)");
fflush(stdin);
another=getche();
}
break;
case '0':
fclose(fp);
exit();
}
}
}

Program to Create, Read & write Using File in C

#include "stdio.h"
main()
{
FILE *fp;
char a[80];
int z;
p:
printf("\n1.Write");
printf("\n2.Read");
printf("\n3.Out");
printf\nEnter U R Choice...");
scanf("%d",&z);
switch(z)
{
case 1:
{
fp = fopen("aa.txt","w");
if(fp==NULL)
{
puts("cant get me");
exit();
}
printf("\nEnter new line...\m");
while(strlen(gets(a))>0)
{
fputs(a,fp);
puts("\n");
}
fclose(fp);
getch();
goto p;
}
}

case 2:
{
fp = fopen("aa.txt","r");
if(fp==NULL)
{
puts("cant get me");
exit();
}
while(fgets(a,79,fp)!=NULL)
printf("%s",a);
fclose(fp);
getch();
goto p;
}
}

Use of OR Operator in C

main()
{
char name;
clrscr();
printf("\n enter 1 letter of your name");
scanf(" %c",&name);
if(name=='r'||name=='R')
{
printf("\n your name is ram");
}
else
printf("\n who are you");
getch();
}

Program to generate a fibonaci number Series in C

main()
{
int n,T ;
clrscr();
printf(" program to generate a fibonaci number");
scanf("d",&n);
fib(n);
return 0;
}
fib( int T )

{
int flast = 1, slast = 0;
int curname;
printf(" febonacci number :\n");
printf("\n");
do
{
curname =flast+slast;
printf("%u\n",curname);
slast=flast;
flast=curname;
T--;
}
while(T>1);
return;
getch();
}


Program to get imagenary Roots

#include"stdio.h"
#include"conio.h"
#include"math.h"
main()
{
int a,b,c;
float r1,r2,v1,v2;
clrscr();
printf("enter value of a,b,c");
scanf("%d %d %d",&a,&b,&c);
v1=(b*b)-(4*a*c);
if(v1<0)
{
printf("%f",v1);
printf("\n the root are imagenary");
}
else
v2=sqrt(v1);
r1=((-b)+v2)/(a*2);
r2=((-b)-v2)/(2*a);
printf("answer=%f\t%f",r1,r2);
getch();
}

Debit Credit Project in C

main()
{
int accno[10],amt[10],de,cr;
int i=1,j,z;

aa:
clrscr();
printf("\n 1. add record...");
printf("\n 2. debit..");
printf("\n 3.credit...");
printf("\n 4.disply..");
printf("\n 5.exit..");
printf("\n enter your choice number ....\n");
scanf("%d",&z);
switch(z)
{
case 1:
{
printf("enter accno & opening amount.....\n");
scanf("%d %d",&accno[i],&amt[i]);
i=i+1;
}
getch();
goto aa;

case 2:
printf("Enter the Accno & debit amount...\n");
scanf("%d %d",&accno,&de);
amt[j]=amt[j]+de;
getch();
goto aa;

case 3:
printf("\nEnter the Accno & credit amount..\n");
scanf("%d %d",&accno,&cr);
amt[j]=amt[j]-cr;
printf("%d",amt[j]);
getch();
goto aa;

case 4:
clrscr();
printf("Accno & Amount....\n");
printf("====================");
for(j=1;j{
printf("\n%d %d",accno[j],amt[j]);
}
getch();
goto aa;

case 5:
break;
}

}

Program to show getpixel() application

#include "graphics.h"

void main()
{
int driver = DETECT,mode;
int i;

initgraph(&driver,&mode,"c:\\borlandc\\bgi");
line(100,100,500,100);
for ( i = 20; i < 300; i++ )
if ( getpixel(300,i) == WHITE )
{
setcolor(BLUE);
circle(300,i,5);
}
else
putpixel(300,i,WHITE);
getch();
closegraph();
}

program to draw a interesting picture using line

#include "graphics.h"
#include "math.h
#include "conio.h"

void main(void)
{
int driver = DETECT,mode;
int x[10],y[10];
int x_center = 360, y_center = 180, rad = 100;
int i,j;

initgraph(&driver,&mode,"c:\\tc\\bgi");
for ( i = 0; i <>
{
x[i] = x_center + rad * cos(36*i*3.14159/180);
y[i] = y_center + rad * sin(36*i*3.14159/180);
}
for ( i = 0; i <>
for ( j = 0; j <>
line(x[i],y[i],x[j],y[j]);
getch(); /* press any key return to TEXT mode */
closegraph();
}