Bug 13842

Summary: Simplify Color::light() and Color::dark()
Product: WebKit Reporter: mitz
Component: Layout and RenderingAssignee: Nobody <webkit-unassigned>
Status: RESOLVED FIXED    
Severity: Trivial    
Priority: P3    
Version: 523.x (Safari 3)   
Hardware: Mac   
OS: OS X 10.4   
Attachments:
Description Flags
Simpler implementation
none
Simpler implementation darin: review+

mitz
Reported 2007-05-23 07:10:16 PDT
The RGB->HSV->RGB conversion Color::light() and Color::dark() do is an overkill. I don't think those functions are hot but what they do can be described and implemented in simpler terms.
Attachments
Simpler implementation (4.28 KB, patch)
2007-05-23 07:11 PDT, mitz
no flags
Simpler implementation (4.40 KB, patch)
2007-05-23 07:40 PDT, mitz
darin: review+
mitz
Comment 1 2007-05-23 07:11:25 PDT
Created attachment 14683 [details] Simpler implementation
mitz
Comment 2 2007-05-23 07:40:01 PDT
Created attachment 14684 [details] Simpler implementation
Darin Adler
Comment 3 2007-05-23 07:57:12 PDT
Comment on attachment 14684 [details] Simpler implementation Does this really give the same values as before? If so, seems fine. But I was under the impression that scaling value didn't affect the 3 channels equally. I don't think dark() properly handles the case where all three of r, g, and b are zero. So review- because of that. Otherwise, would be r=me.
mitz
Comment 4 2007-05-23 09:09:22 PDT
Comment on attachment 14684 [details] Simpler implementation (In reply to comment #3) > (From update of attachment 14684 [details] [edit]) > Does this really give the same values as before? If so, seems fine. But I was > under the impression that scaling value didn't affect the 3 channels equally. If you look at convertHSVToRGB() you'll see that its result can be expressed as "v times some vector x whose components do not depend on v", so indeed for given h and s components, scaling v affects the 3 channels equally. (IIRC in the HSL space, L doesn't affect all channels equally, since L=1.0 maps to white). > I don't think dark() properly handles the case where all three of r, g, and b > are zero. I verified that in that case (v - 0.33f) / v is -inf and therefore the multiplier is 0, so the result is <0,0,0,alpha>, which is what I would expect.
Darin Adler
Comment 5 2007-05-23 09:12:22 PDT
Comment on attachment 14684 [details] Simpler implementation r=me
Mark Rowe (bdash)
Comment 6 2007-05-26 07:00:59 PDT
Landed in r21798.
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