T
Tuco
I am looking for a way to force Windows to flush or invalidate a disk cache
after a CD/DVD has been swapped. (I'm assuming, by the results discussed
below, that Windows uses such a cache.
Background info: I often need to make multiple copies of a CD that our to be
sent out. As part of a verification process, I have begun to include a MD5
checksum file for the files on the CD. (This is a result from others having
problems with some of the copies). Suppose I have more than one copy of a CD.
I have noticed thatverifying the first copy of a CD (using the MD5 sums)
takes X amount of time. Then, when verifying the remaining CDs, the
verification time can be much less than X. I began to wonder if Windows may
not necessarily be reading all the contents of remaining copies, because it
probably thinks that the "same" CD is still in the drive and using the
contents of disk cache when the files were read in from the first copy. Thus,
one may be told that the all the files have been verified when in fact they
haven't been. This problem was confirmed. I had two copies of a DVD. One
copy was good, while the second copy had one file that was bad. Each disk was
verified by rebooting the computer and using MD5 sums for verifying. (The
second disk failed the verification for one file). Then, after rebooting the
computer, copy 1 was verified. Copy 2 was then inserted add passed the
verification process when it should not have.
After some searching, I found a post titled "How to flush the CD-ROM Cache
in Windows XP " (http://bitrazor.com/content/misc/cdromflush.php). I tried
following this procedure when swapping disk in the case I mentioned above
where I know I have one good and one bad copy. However I still got an
incorrect verification stating that copy 2 passed.
So, is there a procedure to force windows to invalidate its CD-ROM cache
when changing CDs.
after a CD/DVD has been swapped. (I'm assuming, by the results discussed
below, that Windows uses such a cache.
Background info: I often need to make multiple copies of a CD that our to be
sent out. As part of a verification process, I have begun to include a MD5
checksum file for the files on the CD. (This is a result from others having
problems with some of the copies). Suppose I have more than one copy of a CD.
I have noticed thatverifying the first copy of a CD (using the MD5 sums)
takes X amount of time. Then, when verifying the remaining CDs, the
verification time can be much less than X. I began to wonder if Windows may
not necessarily be reading all the contents of remaining copies, because it
probably thinks that the "same" CD is still in the drive and using the
contents of disk cache when the files were read in from the first copy. Thus,
one may be told that the all the files have been verified when in fact they
haven't been. This problem was confirmed. I had two copies of a DVD. One
copy was good, while the second copy had one file that was bad. Each disk was
verified by rebooting the computer and using MD5 sums for verifying. (The
second disk failed the verification for one file). Then, after rebooting the
computer, copy 1 was verified. Copy 2 was then inserted add passed the
verification process when it should not have.
After some searching, I found a post titled "How to flush the CD-ROM Cache
in Windows XP " (http://bitrazor.com/content/misc/cdromflush.php). I tried
following this procedure when swapping disk in the case I mentioned above
where I know I have one good and one bad copy. However I still got an
incorrect verification stating that copy 2 passed.
So, is there a procedure to force windows to invalidate its CD-ROM cache
when changing CDs.