[CSS21][ Fixed table layout] Questions on fixed table layout algorithm




In the fixed table layout algorithm, the width of each column is
determined as follows:

    A column element with a value other than 'auto' for the 'width'
property sets the width for that column.
    Otherwise, a cell in the first row with a value other than 'auto' for
the 'width' property determines the width for that column. If the cell
spans more than one column, the width is divided over the columns.
section Fixed table layout

The algorithm given in section
requires as input the set non-auto width of the table (and table borders
in border-collapse: collapse mode or cell spacing in border-collapse:
separate; we'll just assume we have such info) and then one of the
following 2 data:
- one set non-auto width of a column or
- one set non-auto width of a cell in first row

Now, in the table-layout-applies-to-006 testcase, we do not have one of
those 2 data. Because of that, I think the testcase is incorrect. Am I
wrong here?


1- If width of each column is not determinable in a testcase where table
has a set non-auto width and a 'table-layout: fixed' declaration,
shouldn't 'table-layout: auto' algorithm (precision) be triggered and used
Browsers (Firefox 7.0.1, Opera 11.52) seem to disagree with me on this.

2- What is the meaning of an empty colgroup for CSS formatting purposes?
Colgroup can have 0 or more columns. I think an empty colgroup is totally
meaningless and worthless for use in the fixed table layout like in the
following code:

colgroup {width: 100px;}

3- Even if I have

colgroup {width: 400px;}

in a simple 2 column (2 cells per row) table with 'table-layout: fixed',
no one can establish how wide are respectively supposed to be first column
and second column. Am I wrong?

3 testpages on all this:




regards, Gérard
CSS 2.1 Test suite RC6, March 23rd 2011

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Received on Monday, 21 November 2011 00:23:12 UTC