[raymacsor]

plz help me guyzz....pls modify this....this code should have two balls and wen they bump each other they go in separate ways....
i need it as soon as possible guyzz help...thanx

#include<conio.h>
#include<graphics.h>
#include<dos.h>
#include<stdlib.h>
main()
{
 int gdriver=DETECT, gmode, r, col, row, pathx, pathy,speed;
 int forecolor, bkcolor, linecolor;
 initgraph(&gdriver,&gmode,"C:\\TCLITE\\BGI");
 randomize();
 r=random(40)+10;
 col=r+random(640-2*r);
 row=r+random(480-2*r);
 forecolor=random(14)+1;
 bkcolor=random(14)+1;
 linecolor=random(14)+1;
 speed=random(5)+1;
 pathx=(pathy=speed);
 setbkcolor(bkcolor);
 while (!kbhit())
 {
      col+=pathx;
      row+=pathy;
      setcolor(linecolor);
      circle(col,row,r);
      setfillstyle(SOLID_FILL,forecolor);
      floodfill(col,row,linecolor);
      delay(10);
      setfillstyle(SOLID_FILL,bkcolor);
      floodfill(col,row,linecolor);
      setcolor(bkcolor);
      circle(col,row,r);
      col+=pathx;
      row+=pathy;
      if (col+r>640) pathx=-(speed);
      if (row+r>480) pathy=-(speed);
      if (col-r<0) pathx=speed;
      if (row-r<0) pathy=speed;
 }
 closegraph();

}

[shabbir]

I edited the code to put the syntax highlighting syntax. Reading http://www.go4expert.com/showthread.php?t=168 will help you.

Also regarding your query it looks like you are looking for a solution but did not mention where exactly the problem you are facing or what is the nature of the problem. Please clarify on this so that we can help you in this regard.

[raymacsor]

i need a code dat has 2 balls coz this code has only one ball and when the 2 balls bump each other they will go in separate way...

[coderzone]

Quote:

Originally Posted by raymacsor

i need a code dat has 2 balls coz this code has only one ball and when the 2 balls bump each other they will go in separate way...

We cannot give you the complete program for you but if you are stuck at any point then we can help.

[shabbir]

Quote:

Originally Posted by coderzone

We cannot give you the complete program for you but if you are stuck at any point then we can help.

exactly.