C Program Using 8086 Interrupts to Restrict Mouse Pointer Into a Circle of Given Center and Radius. The mouse ponter will be restricted to a circle of user specified center and radius using the interrupt 33 of 8086 in c complier. I have tested this program in Turbo
C compiler.
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Mouse pointer restricted into a circle in Turbo C |
#include<stdio.h>
#include<conio.h>
#include<dos.h>
#include<graphics.h>
void main()
{
int x,y,tx,ty,r;
union REGS inreg, outreg;
/* request auto detection */
int gdriver = DETECT, gmode, errorcode;
clrscr();
/* initialize graphics and local variables */
initgraph(&gdriver, &gmode,"C:\\TC\\BGI\\" );
/* read result of initialization */
errorcode = graphresult();
if (errorcode != grOk) /* an error occurred */
{
printf("Graphics error: %s\n", grapherrormsg(errorcode));
printf("Press any key to halt:");
getch();
exit(1); /* terminate with an error code */
}
setcolor(getmaxcolor());
printf("\nEnter the center of circle. x and y");
scanf("%d%d",&x,&y);
printf("\nEnter the radius of circle");
scanf("%d",&r);
clrscr();
circle(x,y,r);
inreg.x.ax=0x1;
int86(0x33,&inreg,&outreg);
inreg.x.ax=0x7;
inreg.x.cx=x-r;
inreg.x.dx=x+r;
int86(0x33,&inreg,&outreg);
inreg.x.ax=0x8;
inreg.x.cx=y-r;
inreg.x.dx=y+r;
int86(0x33,&inreg,&outreg);
do
{
inreg.x.ax=0x3;
int86(0x33,&inreg,&outreg);
tx=outreg.x.cx;
ty=outreg.x.dx;
if((tx-x)*(tx-x)+(ty-y)*(ty-y)>r*r)
{
if(tx<=x)
{
if(ty<y)
do{tx++; ty++;}while((tx-x)*(tx-x)+(ty-y)*(ty-y)>=r*r);
else
do{tx++; ty--;}while((tx-x)*(tx-x)+(ty-y)*(ty-y)>=r*r);
}
else
{
if(ty<=y)
do{tx--; ty++;}while((tx-x)*(tx-x)+(ty-y)*(ty-y)>=r*r);
else
do{tx--; ty--;}while((tx-x)*(tx-x)+(ty-y)*(ty-y)>=r*r);
}
inreg.x.ax=0x4;
inreg.x.cx=tx;
inreg.x.dx=ty;
int86(0x33,&inreg,&outreg);
}
}while(1);
}
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