C / C++ / MFC

p_1960 wrote:

im planning to start with Programming windows with MFC (Jeff procise Prosise)..

   2x1-x2=1     ...................[1]
-xi+2xi+1=0, i=1,2,.....,n-2. .....[2]
-xn-1+2xn=1  ......................[3]

 n= ? : rem n value should be set

 for j=1 to 50
 rem  50 is max iterations set to.

 for i=1 to n-2
  if i=2 then
   x(2)= 2*x(1)-1
  end if

  x(i)= 2*x(i+1)

 next i
  x(n-1) = 2*x(n) -1
next j

1. It's just a command line, as passed to CMD, so put double quotes around the path as it contains spaces:
   
   

   system("\"c:\\Program Files\\Internet Explorer\\iexplorer.exe\"");
   

2. Isn't it iexplore.exe, not iexplorer.exe?

Xi' = (Xi-Xo)cosθ - (Yi-Yo)sinθ
Yi' = (Xi-Xo)sinθ + (Yi-Yo)cosθ

void rotate (float angle)
{
        ? = (? - px)cos(angle) - (? - py)sin(angle);
	? = (? - px)sin(angle) + (? - py)cos(angle);
}

void rotate (float angle)
{
        rectangleNew[0].x = (rectangle[0].x - px) * cos(angle) - (rectangle[0].y - py) * sin(angle);
	rectangleNew[0].y = (rectangle[0].x - px) * sin(angle) + (rectangle[0].y - py) * cos(angle);
        rectangleNew[1].x = (rectangle[1].x - px) * cos(angle) - (rectangle[1].y - py) * sin(angle);
        rectangleNew[1].y = (rectangle[1].x - px) * sin(angle) + (rectangle[1].y - py) * cos(angle);
        rectangleNew[2].x = (rectangle[2].x - px) * cos(angle) - (rectangle[2].y - py) * sin(angle);
        rectangleNew[2].y = (rectangle[2].x - px) * sin(angle) + (rectangle[2].y - py) * cos(angle);
        rectangleNew[3].x = (rectangle[3].x - px) * cos(angle) - (rectangle[3].y - py) * sin(angle);
        rectangleNew[3].y = (rectangle[3].x - px) * sin(angle) + (rectangle[3].y - py) * cos(angle);
}