What's BitBlt's DSTINVERT supposed to do?

[anonymous]

The description in MSDN states that DSTINVERT "Inverts the destination rectangle." I imagined that this might mean something like "does a 'bitwise not' on the colors" - for example, a pixel that is RGB(255, 0, 10) becomes RGB(0, 255, 245).

However, when I tried a simple little test, what actually happened was that the entire picture went black.

My test was just a simple form in VB6, with a command button "Command1" and a picture box "Picture1", with the box preloaded with some image, and the code:

Private Sub Command1_Click()
BitBlt Picture1.hDC, 0, 0, Picture1.ScaleWidth, Picture1.ScaleHeight, _
0, 0, 0, DSTINVERT
End Sub

The only other code was the declare of BitBlt from the gdi32 lib.

Am I doing something wrong?

[anonymous]

OK, that was incredibly stupid.

I didn't include the declaration for what DSTINVERT meant. I also didn't include "Option Explicit". So when the VB IDE interpreter encountered it, it assumed that DSTINVERT meant 0, which is equivalent to the raster op BLACKNESS.

[Killer42]

OK, that was incredibly stupid.

I didn't include ...

So this one is sorted, then? Gee, that was an easy one! :)


(This is why I always recommend people use Option Explicit. It just confuses things when VB tries to "help out" by creating things for you like this.)