On 04/10/2014 12:11 PM, Tab Atkins Jr. wrote: > Ah, yes, you're right. The point of recalculating is that frozen > items now have a different size than they had before. Try this out: > > # Sum the sizes of all items on the line: > # for frozen items (initially none, see later in this algorithm for > how items are frozen), > # use their outer main size; > # for all other items, > # both flexible and inflexible, > # use their outer <a>flex base size</a>. > # Subtract this sum from the flex container's inner main size. > # This is the free space. Thanks -- I think that makes sense. ~DanielReceived on Thursday, 10 April 2014 19:17:57 UTC
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