Partitioning question

cpulord1

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Sep 25, 2010
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Hi guys.... a simple question really
I noticed that when doing a clean install on Windows 7, using the boot cd/dvd, during the setup, when telling it to "format" a drive, it literally takes a blink of an eye.
During windows 7 setup/install, I split my 1.5 TB HDD and made a 250 GB primary drive(C:), then 500GB(D:), 500GB(E:), then remaining(F:?).
When I get to windows, only C: drive is accessible and formatted. The 2 500GB drives are RAW and require formatting while the remaining space (which I thought would be F:) is unallocated.

My first question being: if I told it to format during the windows install (which I can't remember if I did or not) would that have worked?
Secondly, formatting in windows 7 seems to take a super long time. When you format during the win7 setup, is that a true reformat? (It takes less than a second!??)
Thirdly, why is the last partition unallocated?

Thanks
 
Solution
Windows obviously formats the partition you choose to install it in (otherwise it wouldn't be able to install itself), but the install program won't format any other partitions unless you specifically ask it to.

If you didn't specifically create a partition for the rest of the space then it will be marked as "unallocated".

A "quick" format writes the volume metadata (empty directories, freespace map, master file table, etc.) and nothing else, so it's very fast. A "full" format actually writes to every sector to verify that its usable and places any sectors that couldn't be used into a "bad block table".
Windows obviously formats the partition you choose to install it in (otherwise it wouldn't be able to install itself), but the install program won't format any other partitions unless you specifically ask it to.

If you didn't specifically create a partition for the rest of the space then it will be marked as "unallocated".

A "quick" format writes the volume metadata (empty directories, freespace map, master file table, etc.) and nothing else, so it's very fast. A "full" format actually writes to every sector to verify that its usable and places any sectors that couldn't be used into a "bad block table".
 
Solution