Hi saulgoode,

On Tue, Apr 28, 2009 at 7:00 PM,
<saulgo...@flashingtwelve.brickfilms.com> wrote:
> Quoting David Gowers <00a...@gmail.com>:
>> The eraser currently does change color values, in the case of layers
>> without alpha (it's like using paintbrush or pencil with the
>> background color). Yahvuu's proposition would make sure it never
>> changed color values because there would be no layers without alpha.
> I don't understand the last sentence (perhaps I don't understand Yahvuu's
> proposition correctly). If color values are never changed, the only place
> that erasures would result in an "image background color" (being revealed)
> is on the bottommost layer. Is that what is being proposed?
No, because your reasoning is oversimplified. The above situation
could occur, but it depends on the image content. Some images would
have transparent areas even on the bottom layer (this is common in web
graphics and icons)

> If so, then I would consider the lack of consistency in the tool's behavior
> across layers to be a problem.

Is there an inconsistency to be had? It will behave just the same as
before, really. Only it will never ever change the colors on any
layer, because it will never encounter a layer without alpha that
requires it to paint with BG color instead.

> If it is not so, what determines whether or not erasure results in the RGB
> part of the "image background color" being blended with the layer contents?

The alpha channel of the respective layers. Erasure never results in
the image background color being blended with the layer contents. The
effect on the layer contents is only a change in alpha channel,  just
like it currently is provided your layer has an alpha channel.

Yahvuu's proposition is essentially
a) have a 'virtual layer' always at the bottom of the stack, filled with a color
b) make all layers have an alpha channel

Nothing more.
It does have some ramifications to certain functions (like Cut and Float)
but none really to Eraser (just that the eraser code can be
simplified. most likely)

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