Skip to content
  • Categories
  • Recent
  • Tags
  • Popular
  • World
  • Users
  • Groups
Skins
  • Light
  • Cerulean
  • Cosmo
  • Flatly
  • Journal
  • Litera
  • Lumen
  • Lux
  • Materia
  • Minty
  • Morph
  • Pulse
  • Sandstone
  • Simplex
  • Sketchy
  • Spacelab
  • United
  • Yeti
  • Zephyr
  • Dark
  • Cyborg
  • Darkly
  • Quartz
  • Slate
  • Solar
  • Superhero
  • Vapor

  • Default (No Skin)
  • No Skin
Collapse
Code Project
  1. Home
  2. General Programming
  3. C / C++ / MFC
  4. Rotate rerctangle in Bitmap

Rotate rerctangle in Bitmap

Scheduled Pinned Locked Moved C / C++ / MFC
graphicshardwarehelpquestion
20 Posts 4 Posters 0 Views 1 Watching
  • Oldest to Newest
  • Newest to Oldest
  • Most Votes
Reply
  • Reply as topic
Log in to reply
This topic has been deleted. Only users with topic management privileges can see it.
  • C CPallini

    OK (sorry if I didn't get you). Now, what is the problem with your code (expected behaviour vs observed one)?. :)

    If the Lord God Almighty had consulted me before embarking upon the Creation, I would have recommended something simpler. -- Alfonso the Wise, 13th Century King of Castile.
    This is going on my arrogant assumptions. You may have a superb reason why I'm completely wrong. -- Iain Clarke
    [My articles]

    R Offline
    R Offline
    raju_shiva
    wrote on last edited by
    #11

    Now i want to rotate it for the given degree.How can i do it. I am bit confused please help me Thanks Raj

    C 1 Reply Last reply
    0
    • R raju_shiva

      Now i want to rotate it for the given degree.How can i do it. I am bit confused please help me Thanks Raj

      C Offline
      C Offline
      CPallini
      wrote on last edited by
      #12

      You are defining the rectangle via offsets from center, that is

      P0={-x1,-y2}, P1{-x1, y1}, P2={x2,y1}, P3={x2,-y2}

      hence, if you wan't rotate with angle phi around the center, than you should compute:

      Pr = { x * cos(phi) + y * sin(phi), -x * sin(phi) + y * cos(phi)}

      i.e.:

      P0R= { -x1 * cos(phi) -y2*sin(phi), x1 * sin(phi) - y2 * cos(phi)}
      P1R =...
      P2R =...
      P3R =...

      and then connect the center+PiR points the way you did before. :)

      If the Lord God Almighty had consulted me before embarking upon the Creation, I would have recommended something simpler. -- Alfonso the Wise, 13th Century King of Castile.
      This is going on my arrogant assumptions. You may have a superb reason why I'm completely wrong. -- Iain Clarke
      [My articles]

      R 1 Reply Last reply
      0
      • C CPallini

        You are defining the rectangle via offsets from center, that is

        P0={-x1,-y2}, P1{-x1, y1}, P2={x2,y1}, P3={x2,-y2}

        hence, if you wan't rotate with angle phi around the center, than you should compute:

        Pr = { x * cos(phi) + y * sin(phi), -x * sin(phi) + y * cos(phi)}

        i.e.:

        P0R= { -x1 * cos(phi) -y2*sin(phi), x1 * sin(phi) - y2 * cos(phi)}
        P1R =...
        P2R =...
        P3R =...

        and then connect the center+PiR points the way you did before. :)

        If the Lord God Almighty had consulted me before embarking upon the Creation, I would have recommended something simpler. -- Alfonso the Wise, 13th Century King of Castile.
        This is going on my arrogant assumptions. You may have a superb reason why I'm completely wrong. -- Iain Clarke
        [My articles]

        R Offline
        R Offline
        raju_shiva
        wrote on last edited by
        #13

        CPallini wrote:

        P0R= { -x1 * cos(phi) -y2*sin(phi), x1 * sin(phi) - y2 * cos(phi)}

        I am confused with it??? Now suppose my values are:

        x1 = 330,x2 = 330,y1=330,y2=330
        Origin(center.x,cebter.y) = Origin(525,454);

        I have to do the get the new x1,x2,y1,y2 before calling MoveToEx and LineTo Am i right?? i.e

        Pr = { x * cos(phi) + y * sin(phi), -x * sin(phi) + y * cos(phi)}

        for each x1,x2,y1,y2. Then call the function

        MoveToEx(pCellInfo->hDC,center.x-x1,center.y-y2,NULL); // where x1 = Pr1,x2=Pr2....y2 = Pr4
        LineTo(pCellInfo->hDC,center.x-x1,center.y+y1);
        LineTo(pCellInfo->hDC,center.x+x2,center.y+y1);
        LineTo(pCellInfo->hDC,center.x+x2,center.y-y2);
        LineTo(pCellInfo->hDC,center.x-x1,center.y-y2);

        Thanks Raj

        C 1 Reply Last reply
        0
        • R raju_shiva

          CPallini wrote:

          P0R= { -x1 * cos(phi) -y2*sin(phi), x1 * sin(phi) - y2 * cos(phi)}

          I am confused with it??? Now suppose my values are:

          x1 = 330,x2 = 330,y1=330,y2=330
          Origin(center.x,cebter.y) = Origin(525,454);

          I have to do the get the new x1,x2,y1,y2 before calling MoveToEx and LineTo Am i right?? i.e

          Pr = { x * cos(phi) + y * sin(phi), -x * sin(phi) + y * cos(phi)}

          for each x1,x2,y1,y2. Then call the function

          MoveToEx(pCellInfo->hDC,center.x-x1,center.y-y2,NULL); // where x1 = Pr1,x2=Pr2....y2 = Pr4
          LineTo(pCellInfo->hDC,center.x-x1,center.y+y1);
          LineTo(pCellInfo->hDC,center.x+x2,center.y+y1);
          LineTo(pCellInfo->hDC,center.x+x2,center.y-y2);
          LineTo(pCellInfo->hDC,center.x-x1,center.y-y2);

          Thanks Raj

          C Offline
          C Offline
          CPallini
          wrote on last edited by
          #14

          Something like this

          // rotate around center
          double phi = atan(1.0) * 2/3; // 30 degrees
          int x[4];
          int y[4];
          x[0] = -x1 * cos(phi) - y2 * sin(phi);
          y[0] = x1 * sin(phi) - y2 * cos(phi);

          x[1] = -x1 * cos(phi) + y1 * sin(phi);
          y[1] = x1 * sin(phi) + y1 * cos(phi);

          x[2] = x2 * cos(phi) + y1 * sin(phi);
          y[2] = -x2 * sin(phi) + y1 * cos(phi);

          x[3] = x2 * cos(phi) - y2 * sin(phi);
          y[3] = -x2 * sin(phi) - y2 * cos(phi);

          for (int i=0; i<4; i++)
          {
          x[i] += center.x;
          y[i] += center.y;
          }

          MoveToEx(pCellInfo->hDC,x[3],y[3],NULL);
          for (int i=0; i<4; i++)
          {
          LineTo(pCellInfo->hDC, x[i],y[i]);
          }

          I suppose. :)

          If the Lord God Almighty had consulted me before embarking upon the Creation, I would have recommended something simpler. -- Alfonso the Wise, 13th Century King of Castile.
          This is going on my arrogant assumptions. You may have a superb reason why I'm completely wrong. -- Iain Clarke
          [My articles]

          R R 2 Replies Last reply
          0
          • R raju_shiva

            MoveToEx(pCellInfo->hDC,center.x-x1,center.y-y2,NULL); // Here i will find the new axis from the center axis
            // i.e (center.x point - x1,center.y point-y2)
            //so that i get the start point from where i can start to draw rectangle
            LineTo(pCellInfo->hDC,center.x-x1,center.y+y1); //From here i will draw the rectangle

            I hope you got it,what i am trying to do. If i am doing wrong,please let me know, Thanks for your reply Raj

            P Offline
            P Offline
            Peter_in_2780
            wrote on last edited by
            #15

            When I said "look at the actual values", I didn't mean the expressions, I meant the actual values. In other words, either add some code to print out the values when you call the function or put a breakpoint on the call and inspect the values in your debugger. Then you should be able to figure out what is wrong and work back to where your problem is.

            Software rusts. Simon Stephenson, ca 1994.

            R 1 Reply Last reply
            0
            • C CPallini

              Something like this

              // rotate around center
              double phi = atan(1.0) * 2/3; // 30 degrees
              int x[4];
              int y[4];
              x[0] = -x1 * cos(phi) - y2 * sin(phi);
              y[0] = x1 * sin(phi) - y2 * cos(phi);

              x[1] = -x1 * cos(phi) + y1 * sin(phi);
              y[1] = x1 * sin(phi) + y1 * cos(phi);

              x[2] = x2 * cos(phi) + y1 * sin(phi);
              y[2] = -x2 * sin(phi) + y1 * cos(phi);

              x[3] = x2 * cos(phi) - y2 * sin(phi);
              y[3] = -x2 * sin(phi) - y2 * cos(phi);

              for (int i=0; i<4; i++)
              {
              x[i] += center.x;
              y[i] += center.y;
              }

              MoveToEx(pCellInfo->hDC,x[3],y[3],NULL);
              for (int i=0; i<4; i++)
              {
              LineTo(pCellInfo->hDC, x[i],y[i]);
              }

              I suppose. :)

              If the Lord God Almighty had consulted me before embarking upon the Creation, I would have recommended something simpler. -- Alfonso the Wise, 13th Century King of Castile.
              This is going on my arrogant assumptions. You may have a superb reason why I'm completely wrong. -- Iain Clarke
              [My articles]

              R Offline
              R Offline
              raju_shiva
              wrote on last edited by
              #16

              Thanks a lot ,its working fine. Thanks Raj

              C 1 Reply Last reply
              0
              • P Peter_in_2780

                When I said "look at the actual values", I didn't mean the expressions, I meant the actual values. In other words, either add some code to print out the values when you call the function or put a breakpoint on the call and inspect the values in your debugger. Then you should be able to figure out what is wrong and work back to where your problem is.

                Software rusts. Simon Stephenson, ca 1994.

                R Offline
                R Offline
                raju_shiva
                wrote on last edited by
                #17

                Thank u peter for your reply.Its working fine now Raj

                1 Reply Last reply
                0
                • R raju_shiva

                  Thanks a lot ,its working fine. Thanks Raj

                  C Offline
                  C Offline
                  CPallini
                  wrote on last edited by
                  #18

                  You are welcome. :)

                  If the Lord God Almighty had consulted me before embarking upon the Creation, I would have recommended something simpler. -- Alfonso the Wise, 13th Century King of Castile.
                  This is going on my arrogant assumptions. You may have a superb reason why I'm completely wrong. -- Iain Clarke
                  [My articles]

                  1 Reply Last reply
                  0
                  • C CPallini

                    Something like this

                    // rotate around center
                    double phi = atan(1.0) * 2/3; // 30 degrees
                    int x[4];
                    int y[4];
                    x[0] = -x1 * cos(phi) - y2 * sin(phi);
                    y[0] = x1 * sin(phi) - y2 * cos(phi);

                    x[1] = -x1 * cos(phi) + y1 * sin(phi);
                    y[1] = x1 * sin(phi) + y1 * cos(phi);

                    x[2] = x2 * cos(phi) + y1 * sin(phi);
                    y[2] = -x2 * sin(phi) + y1 * cos(phi);

                    x[3] = x2 * cos(phi) - y2 * sin(phi);
                    y[3] = -x2 * sin(phi) - y2 * cos(phi);

                    for (int i=0; i<4; i++)
                    {
                    x[i] += center.x;
                    y[i] += center.y;
                    }

                    MoveToEx(pCellInfo->hDC,x[3],y[3],NULL);
                    for (int i=0; i<4; i++)
                    {
                    LineTo(pCellInfo->hDC, x[i],y[i]);
                    }

                    I suppose. :)

                    If the Lord God Almighty had consulted me before embarking upon the Creation, I would have recommended something simpler. -- Alfonso the Wise, 13th Century King of Castile.
                    This is going on my arrogant assumptions. You may have a superb reason why I'm completely wrong. -- Iain Clarke
                    [My articles]

                    R Offline
                    R Offline
                    Rick York
                    wrote on last edited by
                    #19

                    The code looks correct but I find it to be slightly offensive to me sensibilities. :) I'm just kidding. I usually make a function or method when I see a sequence of code repeated more than twice. Something like this is generic enough that I definitely would. Of course, it's not your job to put this is into a function. That is an exercise for the reader. ;)

                    C 1 Reply Last reply
                    0
                    • R Rick York

                      The code looks correct but I find it to be slightly offensive to me sensibilities. :) I'm just kidding. I usually make a function or method when I see a sequence of code repeated more than twice. Something like this is generic enough that I definitely would. Of course, it's not your job to put this is into a function. That is an exercise for the reader. ;)

                      C Offline
                      C Offline
                      CPallini
                      wrote on last edited by
                      #20

                      Rick York wrote:

                      he code looks correct but I find it to be slightly offensive to me sensibilities. Smile I'm just kidding. I usually make a function or method when I see a sequence of code repeated more than twice. Something like this is generic enough that I definitely would.

                      Nah, that's childish. You've to set up a linear algebra library, with vectors and rotation matrices, properly overloading the multiplication operator. Of course that's an exercise for you. ;P :laugh: On my defence, I wouldn't write such code repetitions in my own software (well, maybe I do in quick and dirty junk programs), the goal there was being as explicit as possible (the OP had doubts on the steps to take) :)

                      If the Lord God Almighty had consulted me before embarking upon the Creation, I would have recommended something simpler. -- Alfonso the Wise, 13th Century King of Castile.
                      This is going on my arrogant assumptions. You may have a superb reason why I'm completely wrong. -- Iain Clarke
                      [My articles]

                      1 Reply Last reply
                      0
                      Reply
                      • Reply as topic
                      Log in to reply
                      • Oldest to Newest
                      • Newest to Oldest
                      • Most Votes


                      • Login

                      • Don't have an account? Register

                      • Login or register to search.
                      • First post
                        Last post
                      0
                      • Categories
                      • Recent
                      • Tags
                      • Popular
                      • World
                      • Users
                      • Groups