Hi Sandeepmhptr,

Checkout this:-http://www.newlc.com/topic-3539
hopefully you will solve your problem.

Hi sandeepmhptr,

Thanks for advice.I will take care next time onwards.
I hope this might helped you?
Again pasting code below:-
void CMyBitMapClass::CopyFbsBitmapL(CFbsBitmap * orig, CFbsBitmap * dest)
{
TSize s = orig->SizeInPixels();
User::LeaveIfError(dest->Create(s, orig->DisplayMode()));

TBitmapUtil orig_util(orig);
TBitmapUtil dest_util(dest);
orig_util.Begin(TPoint(0,0));
dest_util.Begin(TPoint(0,0), orig_util);

TInt xPos;
for (TInt yPos=0;yPos"<"s.iHeight;yPos++)
{
orig_util.SetPos(TPoint(0,yPos));
dest_util.SetPos(TPoint(0,yPos));
for (xPos=0;xPos {
dest_util.SetPixel(orig_util);
orig_util.IncXPos();
dest_util.IncXPos();
}
}
orig_util.End();
dest_util.End();
}

The above posted code, whilst perfectly correct, it is an extremely slow and roundabout way of copying a bitmap.

If you want an exact copy in a new bitmap (same color format etc) you should use Mem::Copy.

I'd expect it to be somewhere between 10x to 100x as fast:

CFbsBitmap* dstbmp = new (ELeave) CFbsBitmap();
if(KErrNone == dstbmp->Create(aSourceBmp->SizeInPixels(), aSourceBitmap->DisplayMode()))
{
aSourceBmp->LockHeap();

TInt length = aSourceBmp->SizeInPixels().iHeight * aSourceBitmap->DataStride();

Mem::Copy( dstbmp->DataAddress(),  aSourceBmp->DataAddress(),  length);

aSourceBmp->UnlockHeap();
}

Is this use of Mem::Copy the "official" way to do it?

I mean, it works alright, but doesn't this fly in the face of all good object-oriented and abstraction-based programming?

I was surprised anyway to see no method that directly copies a bitmap in the desired way.