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How to create a TIniFile object

Posted on 1998-11-13
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Last Modified: 2010-04-04
Hi.... I'm trying to use a INI file in my program, but I don't know how to initialize it... I have the following procedure that plans to read a file called Monitor.ini:

procedure SetSerial;
var
  Archivo   : TIniFile;
  Puerto    : Byte;
  Baudios   : Integer;
begin
  with Archivo do
    begin
      Create('Monitor.ini');
      Puerto   := ReadInteger('Setup','Puerto',2);
      Baudios  := ReadInteger('Setup','Baud',19200);
      Archivo.Free;
      Emulador.Serial.DeviceName := 'Com' + Chr(Puerto+$30);
      case Baudios of
        2400  : Emulador.Serial.BaudRate := br2400;
        4800  : Emulador.Serial.BaudRate := br4800;
        9600  : Emulador.Serial.BaudRate := br9600;
        19200 : Emulador.Serial.BaudRate := br19200;
      end;
    end;
end;

When I compile it, compiler says the error that Archivo variable might not be initialized. What can I do before calling the Create method in order to initialize the object?

Thank in advance
Jaime
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Question by:skel
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3 Comments
 
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Accepted Solution

by:
sassas081597 earned 150 total points
ID: 1346763
Begin
   Archivo=TIniFile.Create('Monitor.ini');
   with Archivo do
   begin
      Puerto:=...
      ....
   end;
   Archivo.Free;  
end;
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LVL 1

Expert Comment

by:sassas081597
ID: 1346764
Comment to the answer: You should to use the class method TIniFile.Create(...), not the pointer's method Archivo.Create(...).
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LVL 10

Expert Comment

by:viktornet
ID: 1346765
or you could simply do this...

with TIniFile.Create do
try
  //Code....
finally
  Free;
end;

Cheers,
Viktor
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