People forget that the file system itself takes up space -- i want to say that with HFS+ it tends to be about 8.5% of the raw drive capacity. That's ~500MB you lose on an 8GB drive:
(1024*8)*.915 = ~7496
Not sure which file system iPhone uses, but that is definitely a chunk of the missing space. I'm sure the OS and all apps are probably more than 200MB, but who knows how Apple reports that actual utilization in iTunes.
No real mystery here? It is most definitely NOT the difference between MB and MiB.
I just remember that number being used. It could possibly include all the apps, etc, depending on Apple's definition of "free space".
Apparently the UNIX root (/) partition of an 8GB iPod Touch is approximately 300 MB in size. From what I've read, all executable binaries are located in that partition by default, unless you as a hacker do some creative symbolic linking.
The remainder of the storage space in the device is dedicated to a second partition, mounted at /private/var. All media files are stored in that partition.
My guess, and it's purely a guess, is that the space breakdown reported in iTunes deals exclusively with the utilization of the /private/var partition, and doesn't include any executable code at all, in a factory-default configuration.
People forget that the file system itself takes up space -- i want to say that with HFS+ it tends to be about 8.5% of the raw drive capacity. That's ~500MB you lose on an 8GB drive:
(1024*8)*.915 = ~7496
Not sure which file system iPhone uses, but that is definitely a chunk of the missing space. I'm sure the OS and all apps are probably more than 200MB, but who knows how Apple reports that actual utilization in iTunes.
No real mystery here? It is most definitely NOT the difference between MB and MiB.
A file system is taking up 8.5%? Since we know 8GiB is 7.45GB and that iTunes and the iPhone both report that there are 7.3GB capacity, then the file must be taking up about 145MB.
If you can proof that the iPhone is actually larger than 8GB and that the iPhone and iTunes is lying about their capacity I'll entertain it, but the right now the numbers are state the only level proof we have.
A file system is taking up 8.5%? Since we know 8GB is 7.45GiB and that iTunes and the iPhone both report that there are 7.3GiB capacity, then the file must be taking up about 145MiB.
Why would the absolute capacity of the flash chips add up to anything other than 8GiB?
Why would the absolute capacity of the flash chips add up to anything other than 8GiB?
I'm going off on a bit of a tangent here, but it's not actually etched in stone that Flash chips must have powers-of-two for capacities.
It is the most common route. But, it is not universal.
For example, the Atmel AT45 series of dataflash ICs actually are composed of 264-, 528-, or 1056-byte pages.
The AT45DB321 is specified as a 32Mbit part; it actually contains 34603008 bits of storage, or 8192 pages of 528 bytes each.
And solipsim, you were right the first time: 1 GiB has been specified as exactly 1073741284 bytes, and it has been suggested that 1 GB ought to be exactly 1000000000 bytes. So 8 GB (base 10) really would be 7.45 GiB (base 2).
OK, so wouldn't it be most likely that the device has 8GiB, making it 8.59 GB in raw bit capacity, even if the device package says 8GB?
Capacity has always been represented as 10^n while the OS read it as 2^n. Unless flash capacity is marketed as the opposite of HDDs and I'm the last one to the party then it's the same as HDDs.
Capacity has always been represented as 10^n while the OS read it as 2^n. Unless flash capacity is marketed as the opposite of HDDs and I'm the last one to the party then it's the same as HDDs.
I really don't think that makes sense to assume they're the same. Most silicon is made in powers of two, lfmorrison's post on the rare exceptions seems to prove the general rule. If you look up how RAM is sold, it looks like they're still using the unwritten "i".
I really don't think that makes sense to assume they're the same. Most silicon is made in powers of two, lfmorrison's post on the rare exceptions seems to prove the general rule. If you look up how RAM is sold, it looks like they're still using the unwritten "i".
As Palter pointed out, Apple is using 8GB to mean to 8 Billion bytes, so that mystery is resolved. It really is confusing. Even when using binary notation the use of GB vs GiB differs between JEDEC and IEC, according to Wikiepdia.
As Palter pointed out, Apple is using 8GB to mean to 8 Billion bytes, so that mystery is resolved. It really is confusing. Even when using binary notation the use of GB vs GiB differs between JEDEC and IEC, according to Wikiepdia.
What is happening to the extra bits? Most flash ICs aren't made that way. I think Apple is using a fairly standard chip type, so there should be unused, non-user accessible bits if you're right. What is happening to those extra bits?
What is happening to the extra bits? Most flash ICs aren't made that way. I think Apple is using a fairly standard chip type, so there should be unused, non-user accessible bits if you're right. What is happening to those extra bits?
I'm not sure what you mean by that. Is a 1GB USB flash drive actually 1GB using binary notation or actually slightly over 1 Billion bytes?
What is happening to the extra bits? Most flash ICs aren't made that way. I think Apple is using a fairly standard chip type, so there should be unused, non-user accessible bits if you're right. What is happening to those extra bits?
Lots of things could be happening.
A 250 GB hard drive actually has more than 250 billion bytes. Some portions of the disk are reserved by the drive controller (not the filesystem, or any other OS-related thing, mind you - this is one level up from hardware, and no sign of it is ever given to the host CPU) to automatically swap out failed sectors as write failures are detected. Maybe something similar is done by the embedded memory controller inside the Flash chip.
To go along with that, hard drive sectors physically have extra out-of-channel bits attached to them (not directly accessible by the OS, again, this is all handled by the hard drive controller itself) to handle things like ECC signatures, etc. For a Flash chip that has an external interface that is designed to appear to the host computer as though it were an LBA hard drive (such as some of Toshiba's flash chip offerings), something very similar might be at work.
Extra out-of-channel bits might also be useful for use with hardware-driven wear levelling techniques.
It is very possible that some or all of these features might be employed within the chip's embedded memory controlling hardware itself, before any software-driven overhead such as filesystems come into play. They may be completely honest in saying that there are (8 * 2^30) bytes physically present in the IC, but once the embedded memory controller has taken its piece of the action, some of those bytes may not be externally visible.
I'm not sure what you mean by that. Is a 1GB USB flash drive actually 1GB using binary notation or actually slightly over 1 Billion bytes?
I'm saying that the bare chip used should generally have a binary-based quantity. It looks like it's gone when it's packaged into a human useable device, like a phone, SD card or a USB thumb drive. I'm just curious what happened to it.
I'm going off on a bit of a tangent here, but it's not actually etched in stone that Flash chips must have powers-of-two for capacities.
It is the most common route. But, it is not universal.
For example, the Atmel AT45 series of dataflash ICs actually are composed of 264-, 528-, or 1056-byte pages.
The AT45DB321 is specified as a 32Mbit part; it actually contains 34603008 bits of storage, or 8192 pages of 528 bytes each.
And solipsim, you were right the first time: 1 GiB has been specified as exactly 1073741284 bytes, and it has been suggested that 1 GB ought to be exactly 1000000000 bytes. So 8 GB (base 10) really would be 7.45 GiB (base 2).
Since we are dealing with base 2 binary, converting to a base 10 unit system for marketing is pointless, unless of course we manage to extend base 2 boolean logic to include fuzzy logic and self-awareness coding in Artificial Intelligence via critical reasoning gate logic.
I tried out several devices. They all exceed the standard decimal size, but fall short of what I'd expect for a binary size.
4GB iPod Total Capacity t 3.8 GB (4,095,737,856 Bytes) (Which Apple said 1GB = 1 billion bytes in the fine print of their nano page)
128MB CF card: Total Capacity t122.3 MB (128,188,416 Bytes)
256MB CF card: Total Capacity t246.0 MB (257,949,696 Bytes)
512MB CF card: Total Capacity t488.7 MB (512,483,328 Bytes)
2GB CF card Total Capacity t 1.9 GB (2,048,901,120 Bytes)
1GB SD card: Total Capacity t 968.8 MB (1,015,808,000 Bytes)
I'm confused again. Have you not gotten it yet? I checked out the bytes to GiB differences you posted and they are all correct. I knew they would be but I felt I had to check if I wanted to comment on them.
The drawn out way to figure it out is:4,095,737,856 Bytes ÷ 1024 = 3,999,744 KibiBytes ÷ 1024 = 3906 MebiBytes ÷ 1024 = 3.814453125 GibiBytes = 3.8GB (rounded)Or you can just use one of the many byte convertors out there.
I'm confused again. Have you not gotten it yet? I checked out the bytes to GiB differences you posted and they are all correct. I knew they would be but I felt I had to check if I wanted to comment on them.
The drawn out way to figure it out is:4,095,737,856 Bytes ÷ 1024 = 3,999,744 KibiBytes ÷ 1024 = 3906 MebiBytes ÷ 1024 = 3.814453125 GibiBytes = 3.8GB (rounded)
Or you can just use one of the many byte convertors out there.
You are missing what I am saying. Those aren't simple power of two numbers, despite being raw device numbers. They aren't simple power of ten numbers either. 4 GiB is supposed to be 2^32 = 4 294 967 296. 4GB is supposed to be 4x10^9 = 4 000 000 000. What I'm saying is that I don't get why it's not aligning to either system.
You are missing what I am saying. Those aren't simple power of two numbers, despite being raw device numbers. They aren't simple power of ten numbers either. 4 GiB is supposed to be 2^32 = 4 294 967 296. 4GB is supposed to be 4x10^9 = 4 000 000 000. What I'm saying is that I don't get why it's not aligning to either system.
Gotcha.
Take my iTB HDD for example. It's actually 931.5GB or 1,000,204,886,016Bytes. I'm gettting an extra 195.3945 MiB for free, if you want to look at like that. The platters are made up of (from smallest to largest) Sectors, Heads and Cylinders. To make exactly 1 Billion bytes on the drive they would have been waiting space. I'm not even certain it could function without the exact number of bytes in each sector, but I really don't know.
edit: The Wikipage below may more effectual in explaining it:
Comments
(1024*8)*.915 = ~7496
Not sure which file system iPhone uses, but that is definitely a chunk of the missing space. I'm sure the OS and all apps are probably more than 200MB, but who knows how Apple reports that actual utilization in iTunes.
No real mystery here? It is most definitely NOT the difference between MB and MiB.
I just remember that number being used. It could possibly include all the apps, etc, depending on Apple's definition of "free space".
Apparently the UNIX root (/) partition of an 8GB iPod Touch is approximately 300 MB in size. From what I've read, all executable binaries are located in that partition by default, unless you as a hacker do some creative symbolic linking.
The remainder of the storage space in the device is dedicated to a second partition, mounted at /private/var. All media files are stored in that partition.
Source:
http://www.harecoded.com/recovering-...or-ipod-touch/
My guess, and it's purely a guess, is that the space breakdown reported in iTunes deals exclusively with the utilization of the /private/var partition, and doesn't include any executable code at all, in a factory-default configuration.
People forget that the file system itself takes up space -- i want to say that with HFS+ it tends to be about 8.5% of the raw drive capacity. That's ~500MB you lose on an 8GB drive:
(1024*8)*.915 = ~7496
Not sure which file system iPhone uses, but that is definitely a chunk of the missing space. I'm sure the OS and all apps are probably more than 200MB, but who knows how Apple reports that actual utilization in iTunes.
No real mystery here? It is most definitely NOT the difference between MB and MiB.
A file system is taking up 8.5%? Since we know 8GiB is 7.45GB and that iTunes and the iPhone both report that there are 7.3GB capacity, then the file must be taking up about 145MB.
If you can proof that the iPhone is actually larger than 8GB and that the iPhone and iTunes is lying about their capacity I'll entertain it, but the right now the numbers are state the only level proof we have.
A file system is taking up 8.5%? Since we know 8GB is 7.45GiB and that iTunes and the iPhone both report that there are 7.3GiB capacity, then the file must be taking up about 145MiB.
Why would the absolute capacity of the flash chips add up to anything other than 8GiB?
Why would the absolute capacity of the flash chips add up to anything other than 8GiB?
My bad. Reverse my GB and GIB listings.
Why would the absolute capacity of the flash chips add up to anything other than 8GiB?
I'm going off on a bit of a tangent here, but it's not actually etched in stone that Flash chips must have powers-of-two for capacities.
It is the most common route. But, it is not universal.
For example, the Atmel AT45 series of dataflash ICs actually are composed of 264-, 528-, or 1056-byte pages.
The AT45DB321 is specified as a 32Mbit part; it actually contains 34603008 bits of storage, or 8192 pages of 528 bytes each.
And solipsim, you were right the first time: 1 GiB has been specified as exactly 1073741284 bytes, and it has been suggested that 1 GB ought to be exactly 1000000000 bytes. So 8 GB (base 10) really would be 7.45 GiB (base 2).
My bad. Reverse my GB and GIB listings.
OK, so wouldn't it be most likely that the device has 8GiB, making it 8.59 GB in raw bit capacity, even if the device package says 8GB?
OK, so wouldn't it be most likely that the device has 8GiB, making it 8.59 GB in raw bit capacity, even if the device package says 8GB?
No, on Apple's on Tech Specs page for the iPhone, it has the standard footnote which reads that "1 GB is 1 billion bytes".
OK, so wouldn't it be most likely that the device has 8GiB, making it 8.59 GB in raw bit capacity, even if the device package says 8GB?
Capacity has always been represented as 10^n while the OS read it as 2^n. Unless flash capacity is marketed as the opposite of HDDs and I'm the last one to the party then it's the same as HDDs.
Capacity has always been represented as 10^n while the OS read it as 2^n. Unless flash capacity is marketed as the opposite of HDDs and I'm the last one to the party then it's the same as HDDs.
I really don't think that makes sense to assume they're the same. Most silicon is made in powers of two, lfmorrison's post on the rare exceptions seems to prove the general rule. If you look up how RAM is sold, it looks like they're still using the unwritten "i".
I really don't think that makes sense to assume they're the same. Most silicon is made in powers of two, lfmorrison's post on the rare exceptions seems to prove the general rule. If you look up how RAM is sold, it looks like they're still using the unwritten "i".
As Palter pointed out, Apple is using 8GB to mean to 8 Billion bytes, so that mystery is resolved. It really is confusing. Even when using binary notation the use of GB vs GiB differs between JEDEC and IEC, according to Wikiepdia.
As Palter pointed out, Apple is using 8GB to mean to 8 Billion bytes, so that mystery is resolved. It really is confusing. Even when using binary notation the use of GB vs GiB differs between JEDEC and IEC, according to Wikiepdia.
What is happening to the extra bits? Most flash ICs aren't made that way. I think Apple is using a fairly standard chip type, so there should be unused, non-user accessible bits if you're right. What is happening to those extra bits?
What is happening to the extra bits? Most flash ICs aren't made that way. I think Apple is using a fairly standard chip type, so there should be unused, non-user accessible bits if you're right. What is happening to those extra bits?
I'm not sure what you mean by that. Is a 1GB USB flash drive actually 1GB using binary notation or actually slightly over 1 Billion bytes?
What is happening to the extra bits? Most flash ICs aren't made that way. I think Apple is using a fairly standard chip type, so there should be unused, non-user accessible bits if you're right. What is happening to those extra bits?
Lots of things could be happening.
A 250 GB hard drive actually has more than 250 billion bytes. Some portions of the disk are reserved by the drive controller (not the filesystem, or any other OS-related thing, mind you - this is one level up from hardware, and no sign of it is ever given to the host CPU) to automatically swap out failed sectors as write failures are detected. Maybe something similar is done by the embedded memory controller inside the Flash chip.
To go along with that, hard drive sectors physically have extra out-of-channel bits attached to them (not directly accessible by the OS, again, this is all handled by the hard drive controller itself) to handle things like ECC signatures, etc. For a Flash chip that has an external interface that is designed to appear to the host computer as though it were an LBA hard drive (such as some of Toshiba's flash chip offerings), something very similar might be at work.
Extra out-of-channel bits might also be useful for use with hardware-driven wear levelling techniques.
It is very possible that some or all of these features might be employed within the chip's embedded memory controlling hardware itself, before any software-driven overhead such as filesystems come into play. They may be completely honest in saying that there are (8 * 2^30) bytes physically present in the IC, but once the embedded memory controller has taken its piece of the action, some of those bytes may not be externally visible.
I'm not sure what you mean by that. Is a 1GB USB flash drive actually 1GB using binary notation or actually slightly over 1 Billion bytes?
I'm saying that the bare chip used should generally have a binary-based quantity. It looks like it's gone when it's packaged into a human useable device, like a phone, SD card or a USB thumb drive. I'm just curious what happened to it.
I'm going off on a bit of a tangent here, but it's not actually etched in stone that Flash chips must have powers-of-two for capacities.
It is the most common route. But, it is not universal.
For example, the Atmel AT45 series of dataflash ICs actually are composed of 264-, 528-, or 1056-byte pages.
The AT45DB321 is specified as a 32Mbit part; it actually contains 34603008 bits of storage, or 8192 pages of 528 bytes each.
And solipsim, you were right the first time: 1 GiB has been specified as exactly 1073741284 bytes, and it has been suggested that 1 GB ought to be exactly 1000000000 bytes. So 8 GB (base 10) really would be 7.45 GiB (base 2).
Since we are dealing with base 2 binary, converting to a base 10 unit system for marketing is pointless, unless of course we manage to extend base 2 boolean logic to include fuzzy logic and self-awareness coding in Artificial Intelligence via critical reasoning gate logic.
4GB iPod Total Capacity
128MB CF card: Total Capacity
256MB CF card: Total Capacity
512MB CF card: Total Capacity
2GB CF card Total Capacity
1GB SD card: Total Capacity
I tried out several devices. They all exceed the standard decimal size, but fall short of what I'd expect for a binary size.
4GB iPod Total Capacity
128MB CF card: Total Capacity
256MB CF card: Total Capacity
512MB CF card: Total Capacity
2GB CF card Total Capacity
1GB SD card: Total Capacity
I'm confused again. Have you not gotten it yet? I checked out the bytes to GiB differences you posted and they are all correct. I knew they would be but I felt I had to check if I wanted to comment on them.
The drawn out way to figure it out is:4,095,737,856 Bytes ÷ 1024 = 3,999,744 KibiBytes ÷ 1024 = 3906 MebiBytes ÷ 1024 = 3.814453125 GibiBytes = 3.8GB (rounded)Or you can just use one of the many byte convertors out there.
I'm confused again. Have you not gotten it yet? I checked out the bytes to GiB differences you posted and they are all correct. I knew they would be but I felt I had to check if I wanted to comment on them.
The drawn out way to figure it out is:4,095,737,856 Bytes ÷ 1024 = 3,999,744 KibiBytes ÷ 1024 = 3906 MebiBytes ÷ 1024 = 3.814453125 GibiBytes = 3.8GB (rounded)
You are missing what I am saying. Those aren't simple power of two numbers, despite being raw device numbers. They aren't simple power of ten numbers either. 4 GiB is supposed to be 2^32 = 4 294 967 296. 4GB is supposed to be 4x10^9 = 4 000 000 000. What I'm saying is that I don't get why it's not aligning to either system.
You are missing what I am saying. Those aren't simple power of two numbers, despite being raw device numbers. They aren't simple power of ten numbers either. 4 GiB is supposed to be 2^32 = 4 294 967 296. 4GB is supposed to be 4x10^9 = 4 000 000 000. What I'm saying is that I don't get why it's not aligning to either system.
Gotcha.
Take my iTB HDD for example. It's actually 931.5GB or 1,000,204,886,016Bytes. I'm gettting an extra 195.3945 MiB for free, if you want to look at like that. The platters are made up of (from smallest to largest) Sectors, Heads and Cylinders. To make exactly 1 Billion bytes on the drive they would have been waiting space. I'm not even certain it could function without the exact number of bytes in each sector, but I really don't know.
edit: The Wikipage below may more effectual in explaining it: