Question

Send/receive an JPEG image

Asked by: Tango42

Hi,
I am trying to make a image transmission program and a image receiving program on two different computers. The idea is that I load an JPEG image in the transmission program and when I push a send button it should send the JPEG image as a stream to the other computer. I use the NMUDP component on both computers. But I keep getting an error 'access violation at 0x77f60b6f: write of address 0xfd4' when I push the send button... Can anyone see what I am doing wrong... Or does anyone have some source code, that sends and receives an image using UDP.... My procedure is shown below:


procedure TTransmitterForm.btnTransmitClick(Sender: TObject);
var
  MyStream: TMemoryStream;
  Jpg: TJPEGImage;
begin
  NMUDP.RemoteHost := Edit1.Text;
  NMUDP.RemotePort := StrToInt(Edit2.Text);
  //Create a JPEG image and assign the loaded picture
  Jpg := TJPEGImage.Create;
  Jpg.Assign(Image.Picture);
  try
   // Create a stream and save the image to the stream
    MyStream := TMemoryStream.Create;
    Jpg.SaveToStream(MyStream);
    MyStream.Position := 0;
    NMUDP.SendStream(MyStream);
  finally
    MyStream.Free;
    Jpg.Free;
  end;
end;

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Asked On
2003-06-17 at 01:29:22ID20650524
Tags

nmudp

,

send

,

stream

Topics

Delphi Programming

,

Graphics

Participating Experts
6
Points
125
Comments
15

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Answers

 

by: ziolkoPosted on 2003-06-17 at 01:34:24ID: 8738617

>>MyStream.Free I'm not sure but calling free before stream is sent may be a problem.
ziolko.

 

by: jwentingPosted on 2003-06-17 at 02:01:01ID: 8738759

It most likely is. Better close the connection first (after checking if it's done of course) and only then free the memory.
Not sure, but you may have to close the stream as well (don't have Delphi on this computer).

 

by: Tango42Posted on 2003-06-17 at 02:06:49ID: 8738786

I have tryed to remove the line MyStream.Free...But the same error keeps popping up...

 

by: Ferruccio68Posted on 2003-06-17 at 02:17:42ID: 8738849

from the TNMUDP help:

create the Stream before the try/finally...

if you have an OnInvalidHost:
When the OnInvalidHost event is called, the InputQuery function (function in delphi OnInvalidHost example)  is called, allowing the user to input a new host name to replace the invalid one previously entered. If the user clicks the Ok Button, the handled parameter is set to TRUE, and the component attempts it's action again with the new hostname. If the hostname is still invalid, or the cancel button was clicked in the dialog, an exception is raised.

the Delphi Help example:
 procedure TForm1.Button1Click(Sender: TObject);
var
  MyStream: TMemoryStream;
  C: String;
begin
  C := Edit1.Text;
  NMUDP1.RemoteHost := '127.0.0.1';
  NMUDP1.ReportLevel := Status_Basic;
  NMUDP1.RemotePort := 6668;
  MyStream := TMemoryStream.Create;
  try
    MyStream.Write(C[1], Length(C));
    NMUDP1.SendStream(MyStream);
  finally
    MyStream.Free;
  end;
end;

When Button1 is clicked, the text in Edit1 is written into the stream MyStream. The SendStream method is used to send the stream to the remote host, which is 127.0.0.1 (local host) in this instance. Note that MyStream is utilized within a try...finally loop to prevent a memory leak.

 

by: Tango42Posted on 2003-06-17 at 03:02:21ID: 8739065

I have looked at the example, but  I think MyStream.Write only works if it is an integer... I Have tried  to replace the line MyStream.Write(C[1], Length(C)); with MyStream.Write(Jpg[1], Length(Jpg)); But this does not work...

 

by: Ferruccio68Posted on 2003-06-17 at 03:11:28ID: 8739109

of course, that's an example about sending strings...

my post was about the freeing on finally, and that's ok in your source too...the difference is in the stream creation (before the try...finally) and if you're passing some code to the OnInvalidHost event setting Handle to true, but the Host is not valid, this raises an exception...

 

by: geobulPosted on 2003-06-17 at 04:35:50ID: 8739504

Hi,

The code below works fine with very small pictures but not with bigger ones:
There are one NMUDP, one Button and two Image components on a form. Image1 has a small bmp loaded in design-time and Image2 is empty. When you press the button, the picture from Image1 is being transferred via NMUDP and assigned to Image2.
------
unit Unit1;

interface

uses
  Windows, Messages, SysUtils, Classes, Graphics, Controls, Forms, Dialogs,
  StdCtrls, NMUDP, ExtCtrls;

type
  TForm1 = class(TForm)
    Image1: TImage;
    Image2: TImage;
    NMUDP1: TNMUDP;
    Button1: TButton;
    Edit1: TEdit;
    procedure Button1Click(Sender: TObject);
    procedure FormCreate(Sender: TObject);
    procedure NMUDP1DataReceived(Sender: TComponent; NumberBytes: Integer;
      FromIP: String; Port: Integer);
  private
    { Private declarations }
  public
    { Public declarations }
  end;

var
  Form1: TForm1;

implementation

{$R *.DFM}

uses jpeg;

procedure TForm1.FormCreate(Sender: TObject);
begin
  NMUDP1.RemoteHost := '127.0.0.1';
  NMUDP1.ReportLevel := Status_Basic;
  NMUDP1.LocalPort := 6668;
  NMUDP1.RemotePort := 6668;
end;

procedure TForm1.Button1Click(Sender: TObject);
var
 MyStream: TMemoryStream;
 Jpg: TJPEGImage;
begin
 //Create a JPEG image and assign the loaded picture
 Jpg := TJPEGImage.Create;
 Jpg.Assign(Image1.Picture.Graphic);
 try
  // Create a stream and save the image to the stream
   MyStream := TMemoryStream.Create;
   try
     Jpg.SaveToStream(MyStream);
     NMUDP1.SendStream(MyStream);
   finally
     MyStream.Free;
   end;
 finally
   Jpg.Free;
 end;
end;

procedure TForm1.NMUDP1DataReceived(Sender: TComponent;
  NumberBytes: Integer; FromIP: String; Port: Integer);
var
  MyStream: TMemoryStream;
  Jpg: TJPEGImage;
begin
 try
  Jpg := TJPEGImage.Create;
  try
    MyStream := TMemoryStream.Create;
    try
      NMUDP1.ReadStream(MyStream);
      MyStream.Position := 0;
      Jpg.LoadFromStream(MyStream);
      Image2.Picture.Assign(jpg);
    finally
      MyStream.Free;
    end;
  finally
    Jpg.Free;
  end;
 except
 end;
end;

end.
-------
Regards, Geo

 

by: geobulPosted on 2003-06-17 at 04:44:25ID: 8739545

OnDataReceived event should be:

procedure TForm1.NMUDP1DataReceived(Sender: TComponent;
  NumberBytes: Integer; FromIP: String; Port: Integer);
var
  MyStream: TMemoryStream;
  Jpg: TJPEGImage;
begin
  Jpg := TJPEGImage.Create;
  try
    MyStream := TMemoryStream.Create;
    try
      NMUDP1.ReadStream(MyStream);
      Jpg.LoadFromStream(MyStream);
      Image2.Picture.Assign(jpg);
      MyStream.Clear;
    finally
      MyStream.Free;
    end;
  finally
    Jpg.Free;
  end;
end;

Regards, Geo

 

by: Tango42Posted on 2003-06-17 at 05:30:02ID: 8739791

Thanks Geo... I tried my original code with a small bimap and it works just fine... I just don´t see why it won´t work with a larger JPEG image.... I will give it some time to see if anyone can solve it...

 

by: geobulPosted on 2003-06-17 at 06:03:23ID: 8740001

I know that there is a size limit of one UDP packet (can't find  the actual value now but it's definately a small one - less than 65535 bytes). A possible solution could be to cut the stream into small pieces, send them one after another and assemble the picture on receipt.

Regards, Geo

 

by: Tango42Posted on 2003-06-17 at 06:40:17ID: 8740308

Hmm... You´r probably right....

 

by: ginsonicPosted on 2003-06-17 at 09:47:28ID: 8742189

Take a look at http://community.borland.com/article/0,1410,20378,00.html
The article come with a sample project.

 

by: Tango42Posted on 2003-06-18 at 07:43:50ID: 8749664

Anyone have an idea on how to cut the stream into small pieces, send them one after another and assemble the picture on receipt?

 

by: emadatPosted on 2003-06-18 at 08:12:54ID: 8749923

Tango42:

If you are using Indy, then you do not have to worry about the packet size; because it takes care of that for you

Regarding your original error; most likely; you are getting the error because you are not converting the jpg correctly.
here is your original code after adding three more lines to it.


procedure TTransmitterForm.btnTransmitClick(Sender: TObject);
var
 MyStream: TMemoryStream;
 Jpg: TJPEGImage;
begin
 NMUDP.RemoteHost := Edit1.Text;
 NMUDP.RemotePort := StrToInt(Edit2.Text);
 //Create a JPEG image and assign the loaded picture
 Jpg := TJPEGImage.Create;
 Jpg.Assign(Image.Picture);
 try

      JPG.CompressionQuality := 50;  // or whatever you need
      JPG.JPEGNeeded;
      JPG.Compress;

  // Create a stream and save the image to the stream
   MyStream := TMemoryStream.Create;
   Jpg.SaveToStream(MyStream);
   MyStream.Position := 0;
   NMUDP.SendStream(MyStream);
 finally
   MyStream.Free;
   Jpg.Free;
 end;
end;

 

by: Tango42Posted on 2003-06-18 at 08:24:10ID: 8750021

emadat:

I tried to add the three lines, but it did not help.... I will try to see if I can solve the problem with indy....

20120131-EE-VQP-002

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