Li-Ta Lo <[EMAIL PROTECTED]> writes: > On Mon, 2004-03-01 at 14:06, Eric W. Biederman wrote: > > > > There are many was to assign resources on a bus. After some > > experiences with tight memory situations I implemented a near optimal > > solution. The solution is optimal if all of your resources are a > > power of 2 in size. > > > > Basically the code is a loop. For each iteration the > > code finds the largest unassigned resource. Then the resource > > constraints of that resource are considered and padding between > > the previous resources and the current resources are inserted if > > necessary. Then we get into the next iteration. > > > > I still don't understand this. Do you mean that now the resource > allocation is not sequential (in the order of devices been enumerated) > and not continuous (there are gaps in allocated address) ?
Correct. The reason the allocation is not sequential (in the order of devices been enumerated) is to reduce the number of gaps in the allocated addresses. So a thought problem to help put this in perspective. First all resources are a power of 2 in size must be that same power of 2 aligned. So if I have 3 resources A,B,C with the following sizes: A: 8K B: 256M C: 4K Allocating them sequentially would use 756M of address space. After allocating A you need nearly 256M of padding to get B 256M aligned. C takes nearly 256M due to padding. Allocating them largest to smallest (B,A,C) uses just 512M of address space. Because the alignment necessary for A is present at the end of B, and the alignment necessary for A is present at the end of C. Eric > > > The reason this is optimal if all of your resources are a power of > > two in size is because if your previous resource is a larger or equal > > power of two no padding will be needed for the current resource. > > _______________________________________________ Linuxbios mailing list [EMAIL PROTECTED] http://www.clustermatic.org/mailman/listinfo/linuxbios
