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  4. i asked to explain this code, but i think i prefer a mental hospital . not a single description about what he done

i asked to explain this code, but i think i prefer a mental hospital . not a single description about what he done

Scheduled Pinned Locked Moved The Weird and The Wonderful
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  • A AspDotNetDev

    I have no idea either, but the letters listed ("qweadzxc") are arranged on the keyboard as a square. That makes me think it's related to video game controls.

    [WikiLeaks Cablegate Cables]

    K Offline
    K Offline
    kdgupta87
    wrote on last edited by
    #3

    good guess, it is something on image analysis my job is use pointer instead of array, can u understand my pain X| X| . :( :( :( :( :(

    1 Reply Last reply
    0
    • K kdgupta87

      void show()
      {
      int man=5;
      string action = string.Empty;
      for (int j = 1; j < indx-1;j++ )
      {
      for (int k = 1; k < indy-1; k++)
      {
      Console.Write(arr[j,k]+ " ");
      int q = Math.Abs(arr[j - 1, k - 1] - arr[j, k]);
      int w = Math.Abs(arr[j - 1, k] - arr[j, k]);
      int e = Math.Abs(arr[j -1, k +1] - arr[j, k]);
      int a = Math.Abs(arr[j , k - 1] - arr[j, k]);
      int d = Math.Abs(arr[j, k + 1] - arr[j, k]);
      int z = Math.Abs(arr[j + 1, k - 1] - arr[j, k]);
      int x = Math.Abs(arr[j + 1, k ] - arr[j, k]);
      int c = Math.Abs(arr[j + 1, k +1] - arr[j, k]);
      if (q < man && w < man && e < man && a < man && d < man && z < man && x < man && c < man)
      {
      action = string.Empty;
      }
      else {//q=w=e
      if (arr[j - 1, k - 1] == arr[j - 1, k] && arr[j - 1, k] == arr[j + 1, k + 1]) {
      //q=w=e=a=d
      if (arr[j - 1, k] == arr[j, k - 1] && arr[j, k - 1] == arr[j - 1, k])
      {
      //q=w=e=a=d=z=x
      if (arr[j - 1, k] == arr[j + 1, k - 1] && arr[j, k - 1] == arr[j + 1, k])
      {
      action = "c";
      }
      else
      { // q=w=e=a=d=z
      if (arr[j - 1, k] == arr[j + 1, k - 1])
      {
      //x=c
      if (arr[j + 1, k] == arr[j + 1, k + 1]) { action = "xc"; }
      else//x!=c
      { action = "x"; }
      }

                                      else
                                      { //q=w=e=a=d!=z
                                          if (arr\[j + 1, k - 1\] == arr\[j + 1, k + 1\])
                                          { //x=c
                                              if (arr\[j + 1, k\] == arr\[j + 1, k + 1\]) { action = "zxc"; }
      
      R Offline
      R Offline
      RobCroll
      wrote on last edited by
      #4

      There are plenty of comments.

      //q=w=e!a!d

      ;) There's no doubt the companies intellectual property is safe hand there. They have even obfuscated 5 to equal man.

      "You get that on the big jobs."

      K 1 Reply Last reply
      0
      • R RobCroll

        There are plenty of comments.

        //q=w=e!a!d

        ;) There's no doubt the companies intellectual property is safe hand there. They have even obfuscated 5 to equal man.

        "You get that on the big jobs."

        K Offline
        K Offline
        kdgupta87
        wrote on last edited by
        #5

        need comments for these comments

        1 Reply Last reply
        0
        • A AspDotNetDev

          I have no idea either, but the letters listed ("qweadzxc") are arranged on the keyboard as a square. That makes me think it's related to video game controls.

          [WikiLeaks Cablegate Cables]

          C Offline
          C Offline
          Chris Meech
          wrote on last edited by
          #6

          AspDotNetDev wrote:

          it's related to video game controls.

          But only for left handed ones. :)

          Chris Meech I am Canadian. [heard in a local bar] In theory there is no difference between theory and practice. In practice there is. [Yogi Berra] posting about Crystal Reports here is like discussing gay marriage on a catholic church’s website.[Nishant Sivakumar]

          A 1 Reply Last reply
          0
          • K kdgupta87

            void show()
            {
            int man=5;
            string action = string.Empty;
            for (int j = 1; j < indx-1;j++ )
            {
            for (int k = 1; k < indy-1; k++)
            {
            Console.Write(arr[j,k]+ " ");
            int q = Math.Abs(arr[j - 1, k - 1] - arr[j, k]);
            int w = Math.Abs(arr[j - 1, k] - arr[j, k]);
            int e = Math.Abs(arr[j -1, k +1] - arr[j, k]);
            int a = Math.Abs(arr[j , k - 1] - arr[j, k]);
            int d = Math.Abs(arr[j, k + 1] - arr[j, k]);
            int z = Math.Abs(arr[j + 1, k - 1] - arr[j, k]);
            int x = Math.Abs(arr[j + 1, k ] - arr[j, k]);
            int c = Math.Abs(arr[j + 1, k +1] - arr[j, k]);
            if (q < man && w < man && e < man && a < man && d < man && z < man && x < man && c < man)
            {
            action = string.Empty;
            }
            else {//q=w=e
            if (arr[j - 1, k - 1] == arr[j - 1, k] && arr[j - 1, k] == arr[j + 1, k + 1]) {
            //q=w=e=a=d
            if (arr[j - 1, k] == arr[j, k - 1] && arr[j, k - 1] == arr[j - 1, k])
            {
            //q=w=e=a=d=z=x
            if (arr[j - 1, k] == arr[j + 1, k - 1] && arr[j, k - 1] == arr[j + 1, k])
            {
            action = "c";
            }
            else
            { // q=w=e=a=d=z
            if (arr[j - 1, k] == arr[j + 1, k - 1])
            {
            //x=c
            if (arr[j + 1, k] == arr[j + 1, k + 1]) { action = "xc"; }
            else//x!=c
            { action = "x"; }
            }

                                            else
                                            { //q=w=e=a=d!=z
                                                if (arr\[j + 1, k - 1\] == arr\[j + 1, k + 1\])
                                                { //x=c
                                                    if (arr\[j + 1, k\] == arr\[j + 1, k + 1\]) { action = "zxc"; }
            
            _ Offline
            _ Offline
            _Erik_
            wrote on last edited by
            #7

            I would prefer to learn chinese rather than trying to understand this mess.

            P 1 Reply Last reply
            0
            • _ _Erik_

              I would prefer to learn chinese rather than trying to understand this mess.

              P Offline
              P Offline
              phil o
              wrote on last edited by
              #8

              So we're all waiting for your brand new article written in Chinese ;)

              1 Reply Last reply
              0
              • K kdgupta87

                void show()
                {
                int man=5;
                string action = string.Empty;
                for (int j = 1; j < indx-1;j++ )
                {
                for (int k = 1; k < indy-1; k++)
                {
                Console.Write(arr[j,k]+ " ");
                int q = Math.Abs(arr[j - 1, k - 1] - arr[j, k]);
                int w = Math.Abs(arr[j - 1, k] - arr[j, k]);
                int e = Math.Abs(arr[j -1, k +1] - arr[j, k]);
                int a = Math.Abs(arr[j , k - 1] - arr[j, k]);
                int d = Math.Abs(arr[j, k + 1] - arr[j, k]);
                int z = Math.Abs(arr[j + 1, k - 1] - arr[j, k]);
                int x = Math.Abs(arr[j + 1, k ] - arr[j, k]);
                int c = Math.Abs(arr[j + 1, k +1] - arr[j, k]);
                if (q < man && w < man && e < man && a < man && d < man && z < man && x < man && c < man)
                {
                action = string.Empty;
                }
                else {//q=w=e
                if (arr[j - 1, k - 1] == arr[j - 1, k] && arr[j - 1, k] == arr[j + 1, k + 1]) {
                //q=w=e=a=d
                if (arr[j - 1, k] == arr[j, k - 1] && arr[j, k - 1] == arr[j - 1, k])
                {
                //q=w=e=a=d=z=x
                if (arr[j - 1, k] == arr[j + 1, k - 1] && arr[j, k - 1] == arr[j + 1, k])
                {
                action = "c";
                }
                else
                { // q=w=e=a=d=z
                if (arr[j - 1, k] == arr[j + 1, k - 1])
                {
                //x=c
                if (arr[j + 1, k] == arr[j + 1, k + 1]) { action = "xc"; }
                else//x!=c
                { action = "x"; }
                }

                                                else
                                                { //q=w=e=a=d!=z
                                                    if (arr\[j + 1, k - 1\] == arr\[j + 1, k + 1\])
                                                    { //x=c
                                                        if (arr\[j + 1, k\] == arr\[j + 1, k + 1\]) { action = "zxc"; }
                
                K Offline
                K Offline
                Keith Badeau
                wrote on last edited by
                #9

                This guy should work for the government--he seems to be an expert on ciphers and cryptography!

                1 Reply Last reply
                0
                • K kdgupta87

                  void show()
                  {
                  int man=5;
                  string action = string.Empty;
                  for (int j = 1; j < indx-1;j++ )
                  {
                  for (int k = 1; k < indy-1; k++)
                  {
                  Console.Write(arr[j,k]+ " ");
                  int q = Math.Abs(arr[j - 1, k - 1] - arr[j, k]);
                  int w = Math.Abs(arr[j - 1, k] - arr[j, k]);
                  int e = Math.Abs(arr[j -1, k +1] - arr[j, k]);
                  int a = Math.Abs(arr[j , k - 1] - arr[j, k]);
                  int d = Math.Abs(arr[j, k + 1] - arr[j, k]);
                  int z = Math.Abs(arr[j + 1, k - 1] - arr[j, k]);
                  int x = Math.Abs(arr[j + 1, k ] - arr[j, k]);
                  int c = Math.Abs(arr[j + 1, k +1] - arr[j, k]);
                  if (q < man && w < man && e < man && a < man && d < man && z < man && x < man && c < man)
                  {
                  action = string.Empty;
                  }
                  else {//q=w=e
                  if (arr[j - 1, k - 1] == arr[j - 1, k] && arr[j - 1, k] == arr[j + 1, k + 1]) {
                  //q=w=e=a=d
                  if (arr[j - 1, k] == arr[j, k - 1] && arr[j, k - 1] == arr[j - 1, k])
                  {
                  //q=w=e=a=d=z=x
                  if (arr[j - 1, k] == arr[j + 1, k - 1] && arr[j, k - 1] == arr[j + 1, k])
                  {
                  action = "c";
                  }
                  else
                  { // q=w=e=a=d=z
                  if (arr[j - 1, k] == arr[j + 1, k - 1])
                  {
                  //x=c
                  if (arr[j + 1, k] == arr[j + 1, k + 1]) { action = "xc"; }
                  else//x!=c
                  { action = "x"; }
                  }

                                                  else
                                                  { //q=w=e=a=d!=z
                                                      if (arr\[j + 1, k - 1\] == arr\[j + 1, k + 1\])
                                                      { //x=c
                                                          if (arr\[j + 1, k\] == arr\[j + 1, k + 1\]) { action = "zxc"; }
                  
                  T Offline
                  T Offline
                  TorstenH
                  wrote on last edited by
                  #10

                  There is a App key for that -> "Delete" I tried to modify an equal code some weeks ago. As I got behind it, I figured out it didn't make much sense at all. regards Torsten

                  I never finish anyth...

                  1 Reply Last reply
                  0
                  • C Chris Meech

                    AspDotNetDev wrote:

                    it's related to video game controls.

                    But only for left handed ones. :)

                    Chris Meech I am Canadian. [heard in a local bar] In theory there is no difference between theory and practice. In practice there is. [Yogi Berra] posting about Crystal Reports here is like discussing gay marriage on a catholic church’s website.[Nishant Sivakumar]

                    A Offline
                    A Offline
                    Asday
                    wrote on last edited by
                    #11

                    Nah, all vidya nowadays has your right hand on the mouse, and your left hand on WASD. Hell, even full keyboard RTS games like StarCraft II do, apart from the occasional N or L hit.

                    1 Reply Last reply
                    0
                    • K kdgupta87

                      void show()
                      {
                      int man=5;
                      string action = string.Empty;
                      for (int j = 1; j < indx-1;j++ )
                      {
                      for (int k = 1; k < indy-1; k++)
                      {
                      Console.Write(arr[j,k]+ " ");
                      int q = Math.Abs(arr[j - 1, k - 1] - arr[j, k]);
                      int w = Math.Abs(arr[j - 1, k] - arr[j, k]);
                      int e = Math.Abs(arr[j -1, k +1] - arr[j, k]);
                      int a = Math.Abs(arr[j , k - 1] - arr[j, k]);
                      int d = Math.Abs(arr[j, k + 1] - arr[j, k]);
                      int z = Math.Abs(arr[j + 1, k - 1] - arr[j, k]);
                      int x = Math.Abs(arr[j + 1, k ] - arr[j, k]);
                      int c = Math.Abs(arr[j + 1, k +1] - arr[j, k]);
                      if (q < man && w < man && e < man && a < man && d < man && z < man && x < man && c < man)
                      {
                      action = string.Empty;
                      }
                      else {//q=w=e
                      if (arr[j - 1, k - 1] == arr[j - 1, k] && arr[j - 1, k] == arr[j + 1, k + 1]) {
                      //q=w=e=a=d
                      if (arr[j - 1, k] == arr[j, k - 1] && arr[j, k - 1] == arr[j - 1, k])
                      {
                      //q=w=e=a=d=z=x
                      if (arr[j - 1, k] == arr[j + 1, k - 1] && arr[j, k - 1] == arr[j + 1, k])
                      {
                      action = "c";
                      }
                      else
                      { // q=w=e=a=d=z
                      if (arr[j - 1, k] == arr[j + 1, k - 1])
                      {
                      //x=c
                      if (arr[j + 1, k] == arr[j + 1, k + 1]) { action = "xc"; }
                      else//x!=c
                      { action = "x"; }
                      }

                                                      else
                                                      { //q=w=e=a=d!=z
                                                          if (arr\[j + 1, k - 1\] == arr\[j + 1, k + 1\])
                                                          { //x=c
                                                              if (arr\[j + 1, k\] == arr\[j + 1, k + 1\]) { action = "zxc"; }
                      
                      B Offline
                      B Offline
                      BillW33
                      wrote on last edited by
                      #12

                      I used to work with people who liked to write code like that. Why add comments, they just slow down the rate at which such fine code can be written. ;) That is one of the reasons I left that job. :-D

                      Just because the code works, it doesn't mean that it is good code.

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