这已经回答了很多次了。答案是:使用scanline
s 而不是非常慢的Pixels
属性。例子:
function CreateBitmapReallyFast: TBitmap;
const
WHITE: TRGBTriple = (rgbtBlue: 255; rgbtGreen: 255; rgbtRed: 255);
BLACK: TRGBTriple = (rgbtBlue: 0; rgbtGreen: 0; rgbtRed: 0);
var
y: Integer;
scanline: PRGBTriple;
x: Integer;
begin
result := TBitmap.Create;
result.SetSize(1920, 1080);
result.PixelFormat := pf24bit;
for y := 0 to result.Height - 1 do
begin
scanline := result.ScanLine[y];
for x := 0 to result.Width - 1 do
begin
if odd(x) then
scanline^ := WHITE
else
scanline^ := BLACK;
inc(scanline);
end;
end;
end;
更酷:
with scanline^ do
begin
rgbtBlue := Random(255);
rgbtGreen := Random(255);
rgbtRed := Random(255);
end;
尝试一下:
procedure TForm1.FormPaint(Sender: TObject);
var
bm: TBitmap;
begin
bm := CreateBitmapReallyFast;
try
Canvas.Draw(0, 0, bm);
finally
bm.Free;
end;
end;
当然,如果你有一个TRGBTriple
or的(打包的)数组TRGBQuad
,并且位图的像素格式相同,你可以简单地Move
将内存中的数据从数组到位图的扫描线。