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# Delphi DWORD and/or comparison

Posted on 2009-05-08
2,461 Views
Hello!

how to compare dword with AND / OR statment?

example

const
DWORD_1 = \$00060000;
DWORD_2 = \$00F80000;
DWORD_3 = \$00010000;
DWORD_5 = \$0000F000;
DWORD_6 = \$000000FF;
DWORD_7 = \$002000FF;

function CheckDWORD(dwBuff: DWORD): String;
begin
if (dwBuff or DWORD_1) = DWORD_1 then
Result := 'DWORD_1,';

if (dwBuff or DWORD_2) = DWORD_2 then
Result := Result + 'DWORD_2,';

if (dwBuff or DWORD_3) = DWORD_3 then
Result := Result + 'DWORD_3,';

if (dwBuff or DWORD_4) = DWORD_4 then
Result := Result + 'DWORD_4,';

if (dwBuff or DWORD_5) = DWORD_5 then
Result := Result + 'DWORD_5,';

if (dwBuff or DWORD_6) = DWORD_6 then
Result := Result + 'DWORD_6';
end;

var
dwBuff: DWORD;
begin
dwBuff := DWORD_3;
//below OK return only "DWORD_3,"
ShowMessage(CheckDWORD(dwBuff));
//but here
dwBuff := DWORD_2 and DWORD_5;
ShowMessage(CheckDWORD(dwBuff));
//return ALL the statments "DWORD_1, DWORD_2, DWORD_3," etc...
//and was supposed to return only "DWORD2, DWORD5" not ALL :(

How can i do?

Thanks!
Best Regards,
Carlos
0
Question by:cebasso
• 3

LVL 8

Expert Comment

ID: 24340648
you can split the DWord to 2 x Word and compare word by word like :

procedure TForm1.AndButtonClick(Sender: TObject);
var    x_dw, y_dw  : Dword     ;
x_hi,x_lo, y_hi, y_lo  : Word;
highword, lowWord  : Word;
begin

x_dw := \$00060000;
y_dw := \$00060000;

{
//  high byte      //  lower byte
x_hi := hi(x_dw);  x_lo := lo(x_dw);
y_hi := hi(y_dw);  x_lo := lo(y_dw);

ShowMessage( IntToStr(x_hi)+'.'+ IntToStr(X_lo));
}

highword:=x_dw div 65536;
lowword:=x_dw mod 65536;

ShowMessage( IntToStr(highword)+'.'+ IntToStr(lowword));

end;
0

LVL 8

Expert Comment

ID: 24342632

a bit more, splitting the DWORD ....

``````      {
//  high byte      //  lower byte
x_hi := hi(x_dw);  x_lo := lo(x_dw);
y_hi := hi(y_dw);  x_lo := lo(y_dw);

ShowMessage( IntToStr(x_hi)+'.'+ IntToStr(X_lo));
}
``````
0

LVL 8

Assisted Solution

BdLm earned 100 total points
ID: 24342641
that the complete code >

unit Unit_dword;
//
//
//
//  have a look at ->  http://www.drbob42.com/uk-bug/hood-07.htm
//  for http://www.experts-exchange.com/Programming/Languages/Pascal/Delphi/Q_24393689.html#a24342632
//
//

interface
uses
Windows, Messages, SysUtils, Variants, Classes, Graphics, Controls, Forms,
Dialogs, StdCtrls;
type

TForm1 = class(TForm)
AndButton: TButton;
DWORDANDButton: TButton;
procedure AndButtonClick(Sender: TObject);
procedure DWORDANDButtonClick(Sender: TObject);
private
{ Private-Deklarationen }
public
{ Public-Deklarationen }
end;
var
Form1: TForm1;
implementation
{\$R *.dfm}
procedure TForm1.AndButtonClick(Sender: TObject);
var    x_dw, y_dw  : Dword     ;
x_hi,x_lo, y_hi, y_lo  : Word;
highword, lowWord  : Word;
begin

//  some  options ....
x_dw := \$00060000;
y_dw := \$00060000;

//  Hight  / Low Word
highword:=x_dw div 65536;
lowword:=x_dw mod 65536;
ShowMessage( IntToStr(highword)+'.'+ IntToStr(lowword));

//  high byte      //  lower byte
x_hi := hi(x_dw);  x_lo := lo(x_dw);
y_hi := hi(y_dw);  x_lo := lo(y_dw);
ShowMessage( IntToStr(x_hi)+'.'+ IntToStr(X_lo));

end;

function DWORDAND (x,y : DWORD) :  DWORD ;
var     x_dw, y_dw  : Dword     ;
x_hi,x_lo, y_hi, y_lo   : Word;
highword_x, lowWord_x   : Word;
highword_y, lowWord_y   : Word;
highword, lowword       : Word;
begin
highword_x:=x div 65536;
lowword_x:=x mod 65536;
highword_y:=y div 65536;
lowword_y:=y mod 65536;

HighWord :=  HighWord_x AND HighWord_y ;
lowword  := lowword_y  AND lowword_y;

result := makeLOng ( highword, lowword) ;

end;

procedure TForm1.DWORDANDButtonClick(Sender: TObject);
begin

ShowMessage (IntToStr(DWORDAND(\$00060000, \$00600000)));
end;
end.
0

LVL 4

Accepted Solution

cebasso earned 0 total points
ID: 24353841
Hello,
@BdLm
Nice method but doesnt work was i needed...
I found a way...
Peter Below wrote:

"The binary OR operator will combine different values to a result that
has all bits set that are set in any of the values.

You are using the wrong operator. If you want to test if the bits set
in DWORD_1 are also set in dwBuff you have to use the AND operator, not
OR:
if (dwBuff AND DWORD_1) = DWORD_1 then
....
The binary AND operator returns a result that has all bits set that are
set in both operants.
Note that dwBuff can still have other bits set that are 0 in DWORD_1,
the test will still succeed. All you test here is that all bits that
are 1 in DWORD_1 are also 1 in dwBuff.
"
Then the solution is use the OR operator to set the values and AND to compare...
Example
dwBuff := DWORD_1 or DWORD5;
and compare with AND
if (dwBuff and DWORD5 = DWORD_5) then
...
Thanks anyway!
Best Regards,
Carlos
0

Expert Comment

ID: 24382766
maybe this can help
------------------------------------------------------------------------------------------------------------------------

unit Unit1;

interface

uses
Windows, Messages, SysUtils, Variants, Classes, Graphics, Controls, Forms,
Dialogs;

type
TForm1 = class(TForm)
procedure FormCreate(Sender: TObject);
private
{ Private declarations }
public
{ Public declarations }
function CheckDWORD(dwBuff: string): String;

end;

var
Form1: TForm1;

const
DWORD_1 = \$00060000;
DWORD_2 = \$00F80000;
DWORD_3 = \$00010000;
DWORD_5 = \$0000F000;
DWORD_6 = \$000000FF;
DWORD_7 = \$002000FF;

implementation

{\$R *.dfm}

function TForm1.CheckDWORD(dwBuff: string): String;
var
sl:TStringList;
I: Integer;
begin
sl:=TStringList.Create;
Result := '';
sl.CommaText := dwBuff;
for I := 0 to sl.Count-1 do
begin
case StrToInt(sl[i]) of
DWORD_1:Result := Result + 'DWORD_1,';
DWORD_2:Result := Result + 'DWORD_2,';
DWORD_3:Result := Result + 'DWORD_3,';
DWORD_5:Result := Result + 'DWORD_5,';
DWORD_6:Result := Result + 'DWORD_6,';
DWORD_7:Result := Result + 'DWORD_7,';
end;
end;
Result := copy(Result,1,length(Result)-1);
sl.free;
end;

procedure TForm1.FormCreate(Sender: TObject);
var
s:string;
begin
s := IntToStr(DWORD_3);
ShowMessage(CheckDWORD(s));
s := IntToStr(DWORD_2)+','+IntToStr(DWORD_5);
ShowMessage(CheckDWORD(s));
end;

end.
0

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