Mega Code Archive
Bitmap crossfade (blending two images in several steps)
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Title: Bitmap crossfade (blending two images in several steps)
Question: How to show an image changing into another.
Answer:
In a previous post I provided a function that takes 2 TBitmap arguments and returns a crossfaded image obtained by combining the two bitmaps.
That can be used to make an animation that shows a bitmap morphing into another bitmap.
unit Unit1;
interface
uses
Windows, Messages, SysUtils, Variants, Classes, Graphics, Controls, Forms,
Dialogs, StdCtrls, ExtCtrls, jpeg;
type
TForm1 = class(TForm)
Button1: TButton;
Image3: TImage;
esteps: TEdit;
procedure Button1Click(Sender: TObject);
private
{ Private declarations }
public
{ Public declarations }
end;
var
Form1: TForm1;
implementation
{$R *.dfm}
function mix(img1, img2:TBitmap; steps, stepno:integer):TBitmap;
var
x, y, height, width:integer;
pimg1, pimg2, pres:PByteArray;
res:TBitmap;
begin
res := TBitmap.Create;
if img1.Height img2.Height then height := img2.Height
else height := img1.Height;
if img1.Width img1.Width then width := img2.Width
else width := img1.Width;
img1.PixelFormat := pf24bit;
img2.PixelFormat := pf24bit;
res.PixelFormat := pf24bit;
res.Width := width;
res.Height := height;
for y := 0 to height - 1 do begin
pimg1 := img1.ScanLine[y];
pimg2 := img2.ScanLine[y];
pres := res.ScanLine[y];
for x := 0 to width - 1 do begin
pres^[x*3] := pimg1^[x*3]*(steps-stepno) div steps +
pimg2[x*3]*stepno div steps;
pres^[x*3+1] := pimg1^[x*3+1]*(steps-stepno) div steps +
pimg2[x*3+1]*stepno div steps;
pres^[x*3+2] := pimg1^[x*3+2]*(steps-stepno) div steps +
pimg2[x*3+2]*stepno div steps;
end;
end;
result := res;
end;
procedure TForm1.Button1Click(Sender: TObject);
var
steps,i:integer;
img1, img2:TBitmap;
begin
if esteps.text '' then begin
steps := strtoint(esteps.text);
img1 := TBitmap.Create;
img2 := TBitmap.Create;
img1.LoadFromFile('e:\untitled.bmp');
img2.LoadFromFile('e:\untitled1.bmp');
for i:=1 to steps do begin
image3.Picture.Bitmap := mix(img1, img2, steps, i);
image3.Refresh;
sleep(30);
end;
img1.Free;
img2.Free;
end
else
messagedlg('Please provide the number of steps for morphing', mterror, [mbok], 0);
end;
end.
I have a form with a button on it, a TImage object and an edit box. The editbox is used to provide the number of steps in wich morphing will occur. I load two bitmaps and the morphing is showed in the TImage object. The function mix is mixing two images in a certain proportion calculated from the values of the steps and stepno.
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