首页 > 代码库 > 取出资源文件中的bitmap,并将其保存到TMemoryStream中,从资源里载入图象而不丢失调色板
取出资源文件中的bitmap,并将其保存到TMemoryStream中,从资源里载入图象而不丢失调色板
从资源里载入图象而不丢失调色板
procedure loadgraphic(naam:string);
var
{ I‘ve moved these in here, so they exist only during the lifetime of the
procedure. }
HResInfo: THandle;
BMF: TBitmapFileHeader;
MemHandle: THandle;
Stream: TMemoryStream;
ResPtr: PByte;
ResSize: Longint;
null:array [0..8] of char;
begin
{ In this first part, you are retrieving the bitmap from the resource.
The bitmap that you retrieve is almost, but not quite, the same as a
.BMP file (complete with palette information). }
strpcopy (null, naam);
HResInfo := FindResource(HInstance, null, RT_Bitmap);
ResSize := SizeofResource(HInstance, HResInfo);
MemHandle := LoadResource(HInstance, HResInfo);
ResPtr := LockResource(MemHandle);
{ Think of a MemoryStream almost as a "File" that exists in memory.
With a Stream, you can treat either the same way! }
Stream := TMemoryStream.Create;
try
Stream.SetSize(ResSize + SizeOf(BMF));
{ Next, you effectively create a .BMP file in memory by first writing
the header (missing from the resource, so you add it)... }
BMF.bfType := $4D42;
Stream.Write(BMF, SizeOf(BMF));
{ Then the data from the resource. Now the stream contains a .BMP file }
Stream.Write(ResPtr^, ResSize);
{ So you point to the beginning of the stream... }
Stream.Seek(0, 0);
{ ...and let Delphi‘s TBitmap load it in }
Bitmap:=tbitmap.create;
Bitmap.LoadFromStream(Stream);
{ At this point, you are done with the stream and the resource. }
finally
Stream.Free;
end;
FreeResource(MemHandle);
end;
取出资源文件中的bitmap,并将其保存到TMemoryStream中,从资源里载入图象而不丢失调色板