VERIFIED FIXED 5573
Bidi algorithm: AN after L has incorrect resolved level 
https://bugs.webkit.org/show_bug.cgi?id=5573
Summary Bidi algorithm: AN after L has incorrect resolved level
mitz
Reported 2005-10-31 10:59:35 PST
Summary: if the embedding level is even, then in the sequence L AN, the AN's resolved level should be greater (by 2) than the L's. WebKit treats them as if they have the same resolved level. (3.3.5 in the UBA). To reproduce: open the test case in Safari. Expected: the two lines to be the same, from left to right: English, Hebrew, Arabic digits. Firefox renders as expected. Actual: on the second line, from left to right: English, Arabic digits, Hebrew.
Attachments
Testcase (442 bytes, text/html)
2005-10-31 11:02 PST, mitz 		no flags
Details
Terminate L run when encountring AN on an even level (716 bytes, patch)
2005-10-31 11:40 PST, mitz 		no flags
DetailsFormatted DiffDiff
Terminate L run when encountring AN on an even level (691 bytes, patch)
2005-10-31 11:43 PST, mitz 		darin: review+
DetailsFormatted DiffDiff
Show Obsolete (1) View All Add attachment proposed patch, testcase, etc.
mitz
Comment 1 2005-10-31 11:02:40 PST
Created attachment 4543 [details] Testcase
mitz
Comment 2 2005-10-31 11:40:16 PST
Created attachment 4544 [details] Terminate L run when encountring AN on an even level
mitz
Comment 3 2005-10-31 11:43:59 PST
Created attachment 4545 [details] Terminate L run when encountring AN on an even level Oops! Corrected style.
mitz
Comment 4 2005-10-31 11:44:24 PST
Comment on attachment 4545 [details] Terminate L run when encountring AN on an even level Fixes the testcase and doesn't break any of the layout tests.
Darin Adler
Comment 5 2005-10-31 14:17:40 PST
Comment on attachment 4545 [details] Terminate L run when encountring AN on an even level In the future, it's probably worthwhile adding a comment in code that has a case that falls through to the next case, since the mistake of leaving out a break is so common and we don't want anyone to think this is a mistake. But that's a minor quibble. r=me
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