I’m just going by what Apple says on its website (I provided the link in post 15 in this thread), and according to the website, I have it the right way around! 
From that link:
The underlined section is incorrect when applied to manufacturers of flash storage devices.
Whatever apples says on any web page, if you are using flash it will have a size of some multiple of a power of two.
Even apple, big as they are are not going to get any chip manufacturer to do something as surreal as produce chips with a size that isn’t a power of two (or multiple thereof).
Of course, there’s nothing that says that apple have to use all the memory on a chip so I suppose that they could simply not use any above 7,999,999,999.
Nonetheless, the amount of physical flash in an 8GB flash based player will be 8,589,934,592 bytes.
To be more precise, the amount of physical flash in a flash based player advertised as containing 8 GB (or 8 GiB) can only be determined by cracking open the case and looking inside.
In any case, Apple doesn’t advertise that the player has an 8 GiB (binary) capacity, it advertises an 8 GB (decimal) capacity. I can’t find any article that describes how much of the flash memory of an iPod touch is available for the operating system + applications. So my calculations will assume at most 8 GB (decimal) available for operating system + applications. If Apple is only using 8 GB (decimal) of the flash memory, obviously the difference between 8 GiB (binary) and 8 GB (decimal) should not be added to the size of the operating system.
Nope.
If it’s an 8GB flash player it will contain 8,589,934,592 bytes of flash.
No one manufactures any semiconductor memory based upon powers of ten. It’s inherent in the way it’s made that it will be based on powers of 2 (and usually 4). For confirmation see here (look under ‘Capacity’.)
BTW, opening the case won’t help. You’d need to break open the chip and use a very powerful microscope. Even then, good luck with counting the bits. 
You miss my point. A player with an advertised capacity of 8 GB can have 4 GiB (if my advertising is a pack of lies.) It can have 4 GiB + 2 GiB + 1 GiB + 512 MiB = 8,053,063,680 bytes = approx. 8 GB (decimal). It can have any combination. Though it is most likely that it will have 8 GiB.
Regardless, the OP’s question was about how much usable storage space she can expect from her iPod Touch. It is reasonable to use, as a starting point, the manufacturer’s advertised usable storage space.
But ram usage doesn’t really tell you much about the size of an operating system. Theoretically you could have a very large OS that only requires a very small amount of ram to operate, the trade off would be execution time as it would have to page things (e.g. the os code itself, data, etc.) in and out much much more than if the entire os was loaded into ram.
Well, if you want to jump into the realms of the ridiculous just so that you can say you were right, go ahead. The chips in the ipod touch are, nonetheless, 8,589,934,592 bytes in size. (See below.)
Except that that starting point is, demonstrably, wrong. See Here
As I said before, do you really think that apple would make such a cock-up of the design that a media player OS needed to page? Certainly it may load different CODECS as needed but it’s not going to be paging normal program code.
qpw3141 -
You said:
This demonstrates a fundamental misunderstanding of how computers work.
One the one hand, you say that using 1.2GB for the OS (on disc) is “disgustingly bloated”, and on the other hand you turn around and say that Windows 7 will fit in 1GB (of RAM). How do you reconcile these two completely incongruous statements?
Or, better yet, try to install Windows 7 on a machine with 1GB of disc.
The iPhone/iPod Touch/iPad has a web browser, media player, photo gallery, games and user installed applications (running in a sandbox VM). In only 128Mb, even single tasking, it swaps. New generation devices with 256Mb running OS 4 can multi task apps, real paging/swap territory. Thus the large OS.
Si
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Why thank you! I love being right. It is theoretically possible for a device, using Flash memory, to have 8 GB (decimal) of memory, approximately, if it has 7.5 GiB (binary) of flash memory.
To prove that the iPod touch actually has 8 GiB of flash memory, you shouldn’t post a link saying that flash memory is measured in binary (which no one is disputing), you should post a link showing the inside of an iPod touch, and showing that it has two 4GiB flash memory chips. Which it seems to have, by the way, according to the pictures I’ve seen of the insides of an iPod touch.
This is another quote showing the difference between binary units and decimal units. I am well aware of the difference. However, to know how much memory you can expect to use with a device, it doesn’t matter how much is physically installed. It matters how much can actually be used. Which is the point I’m making. It is possible that some of the 8 GiB of flash memory is unusable (for operating system + applications.)
Example: I could install 4 GiB of RAM in a system running Windows XP. That doesn’t mean that Windows XP will have access to 4 GiB of RAM. Some of the RAM will never be used. So if I tried to measure how much RAM is being used by the operating system alone, it would be a mistake to say: “With no programs running, windows shows that I have 3 GiB of RAM available. Therefore the OS itself uses up 4 - 3 = 1 GiB of RAM.” That is wrong, because some of those 4 GiB are never used in the first place.
Among other problems I think we may all be using different terms. A very large OS may have a small kernel. The rest of the OS could be the various bits of OS services and pre-installed applications. It is certainly debatable as to whether bits like ftpd and bash are part of the Linux “operating system”.
You keep saying that but it’s just meaningless trash talk.
Well, it’s actually very easy if you know anything about how computers work, 
You seem to want to keep saying this based only on your misunderstanding of the difference between a working memory image and a complete installation.
Windows is a full blown OS that is enormously configurable and works on a vast array of hardware to which can be attached a similarly vast array of peripherals and supports a mind numbing large range of software.
The ipod just has to run one app at a time and only has to do it on one device. The only peripherals it has to attach are a pair of headphones.
It’s not difficult to see why one would take up a lot more room than the other.
That actually works very nicely provided you don’t try and do anything memory intensive with it. You can certainly run a media player in that space without paging.
Well, I said that to keep you happy. Of course, it couldn’t really because there would be no where to fit all those chips.
Anyone with the slightest intelligence can infer one from the other because the ipod is a miniature device and even the most inept of manufacturers isn’t going to do anything so surreal as to synthesize 8 decimal GB of memory by the far more expensive and space and power consuming technique of using 4 times the number of memory chips. :rolleyes:
Er, yes, that is the whole point of the thread.
Did you miss all the other posts?
You’re off on a completely different tack now.
That is what the entire thread is about: Why is so much or the flash memory unavailable. We know that it must be being used for something else.
I think this actually answers the question quite well.
They’ve steadily increased the capabilities of the device and with it, naturally, the various memory requirements.
If you just want a music player that will leave you as much space as possible for music, the ipod touch is not the correct device.
Wrong again: you need to go back to school.
I said 1GB of DISC SPACE.
Yeah, congratulations, you got me. :rolleyes:
I really didn’t think anyone would be so animal stupid as to talk about trying to put W7 on 1GB of disk so I stopped reading at 1GB and assumed RAM.
My bad. Make of that it you will. :dubious:
Wasn’t that answer offered on the first page?!