Derived Licensing Issues (LGPL)
Posted: Sun Oct 21, 2007 3:52 pm
I'm writing program X. However, I run into a small wall. I look to library Y (Which is LGPL.) for guidance. They use a mathematical formula. (let's say x_b-(a/a)+1) I discover said formula from their code, and use it in my code. I don't actually use any of their code, just the formula. (In fact, I do it in a very different way.)
Does this count as derived work? I don't believe it does, because for example, if it did, everything that decoded a PNG would be at the mercy of the license of libpng.
Am I right in my assumptions?
Thanks.
By the by, these are the pieces of code:
This is from SDL_ttf:
And this is mine:
(Sorry for the formatting.)
Does this count as derived work? I don't believe it does, because for example, if it did, everything that decoded a PNG would be at the mercy of the license of libpng.
Am I right in my assumptions?
Thanks.
By the by, these are the pieces of code:
This is from SDL_ttf:
Code: Select all
rdiff = fg.r - bg.r;
gdiff = fg.g - bg.g;
bdiff = fg.b - bg.b;
for( index = 0; index < NUM_GRAYS; ++index ) {
palette->colors[index].r = bg.r + (index*rdiff) / (NUM_GRAYS-1);
palette->colors[index].g = bg.g + (index*gdiff) / (NUM_GRAYS-1);
palette->colors[index].b = bg.b + (index*bdiff) / (NUM_GRAYS-1);
}
Code: Select all
SetForeground(Color(background.color.red+(b.image[_h][_w]*(foreground.color.red-background.color.red)/b.grays),
background.color.green+(b.image[_h][_w]*(foreground.color.green-background.color.green)/b.grays),
background.color.blue+(b.image[_h][_w]*(foreground.color.blue-background.color.blue)/b.grays)));