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Last computer to not use bytes
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I am old enough to remember computers that were not byte oriented. E.g. the first that I used was an ICL 4120. It had 24 bit words which were, when necessary, divided into four 6-bit characters. There were operations to support extracting the 6 bit characters from the words. There were no corresponding operations to extract three 8 bit sub-units.
Today, I am unaware of a computer that is not byte oriented.
Do any remain in production? If not which was the last? Two interpretations of last are interesting: the last to be launched and the last to remain on sale (as new).
Update 1: it seems that some non-byte chips are still in production. I will need to think of a way to narrow down the question.
Update 2: On byte oriented, I did not intend to exclude 16, 32, and 64 bit processors but most popular current CPUs still have substantial support for operations involving 8 bit bytes. I was looking for examples where 8 bits is not significant e.g. my old ICL example, 24 bit words, some support for 6 bit sub-units but no support for 8 bit operations. For example, their memory would not be measured in 8 bit bytes.
memory
asked 4 hours ago
badjohn
4839
 |Â
show 8 more comments
up vote
2
down vote
favorite
I am old enough to remember computers that were not byte oriented. E.g. the first that I used was an ICL 4120. It had 24 bit words which were, when necessary, divided into four 6-bit characters. There were operations to support extracting the 6 bit characters from the words. There were no corresponding operations to extract three 8 bit sub-units.
Today, I am unaware of a computer that is not byte oriented.
Do any remain in production? If not which was the last? Two interpretations of last are interesting: the last to be launched and the last to remain on sale (as new).
Update 1: it seems that some non-byte chips are still in production. I will need to think of a way to narrow down the question.
Update 2: On byte oriented, I did not intend to exclude 16, 32, and 64 bit processors but most popular current CPUs still have substantial support for operations involving 8 bit bytes. I was looking for examples where 8 bits is not significant e.g. my old ICL example, 24 bit words, some support for 6 bit sub-units but no support for 8 bit operations. For example, their memory would not be measured in 8 bit bytes.
memory
asked 4 hours ago
badjohn
4839
1
There are processors today which have "weirdo" sizes, for example PIC16 has 14-bit program memory.
– Wilson
4 hours ago
1
Oh and at least some modern-day Elbrus chips also have a BESM-6 compatibility mode, which is a 48-bit computer which does not have bytes.
– Wilson
4 hours ago
3
A lot of modern DSPs (e.g. TI TMS 32000 series) use "bytes" that have 16 bits: processors.wiki.ti.com/index.php/…
– tofro
4 hours ago
2
Many Harvard type CPUs use different size for programm and data memory, so PICs can be had in 12, 14 and 16 bit program word size. If your question is about Von Neumann machines, then we need to seperate between (logic) byte addressing and physical interface - for example modern x86 CPUs have a physical interface of 8 or more bytes wide, while on a locical level they operate bytewise. and so on. There's no real answer to that.
– Raffzahn
4 hours ago
1
No modern general purpose computer is "byte oriented" (that's a bit of a nebulous term). For example, in the x86_64 architecture, data comes in 8 byte chunks (with each byte being 8 bits). Maybe you should focus on machines that are byte addressable.
– JeremyP
3 hours ago
 |Â
show 8 more comments
up vote
2
down vote
favorite
up vote
2
down vote
favorite
I am old enough to remember computers that were not byte oriented. E.g. the first that I used was an ICL 4120. It had 24 bit words which were, when necessary, divided into four 6-bit characters. There were operations to support extracting the 6 bit characters from the words. There were no corresponding operations to extract three 8 bit sub-units.
Today, I am unaware of a computer that is not byte oriented.
Do any remain in production? If not which was the last? Two interpretations of last are interesting: the last to be launched and the last to remain on sale (as new).
Update 1: it seems that some non-byte chips are still in production. I will need to think of a way to narrow down the question.
Update 2: On byte oriented, I did not intend to exclude 16, 32, and 64 bit processors but most popular current CPUs still have substantial support for operations involving 8 bit bytes. I was looking for examples where 8 bits is not significant e.g. my old ICL example, 24 bit words, some support for 6 bit sub-units but no support for 8 bit operations. For example, their memory would not be measured in 8 bit bytes.
memory
asked 4 hours ago
badjohn
4839
I am old enough to remember computers that were not byte oriented. E.g. the first that I used was an ICL 4120. It had 24 bit words which were, when necessary, divided into four 6-bit characters. There were operations to support extracting the 6 bit characters from the words. There were no corresponding operations to extract three 8 bit sub-units.
Today, I am unaware of a computer that is not byte oriented.
Do any remain in production? If not which was the last? Two interpretations of last are interesting: the last to be launched and the last to remain on sale (as new).
Update 1: it seems that some non-byte chips are still in production. I will need to think of a way to narrow down the question.
Update 2: On byte oriented, I did not intend to exclude 16, 32, and 64 bit processors but most popular current CPUs still have substantial support for operations involving 8 bit bytes. I was looking for examples where 8 bits is not significant e.g. my old ICL example, 24 bit words, some support for 6 bit sub-units but no support for 8 bit operations. For example, their memory would not be measured in 8 bit bytes.
memory
memory
asked 4 hours ago
badjohn
4839
asked 4 hours ago
badjohn
4839
edited 1 hour ago
asked 4 hours ago
badjohn
4839
asked 4 hours ago
badjohn
4839
asked 4 hours ago
badjohn
4839
4839
1
There are processors today which have "weirdo" sizes, for example PIC16 has 14-bit program memory.
– Wilson
4 hours ago
1
Oh and at least some modern-day Elbrus chips also have a BESM-6 compatibility mode, which is a 48-bit computer which does not have bytes.
– Wilson
4 hours ago
3
A lot of modern DSPs (e.g. TI TMS 32000 series) use "bytes" that have 16 bits: processors.wiki.ti.com/index.php/…
– tofro
4 hours ago
2
Many Harvard type CPUs use different size for programm and data memory, so PICs can be had in 12, 14 and 16 bit program word size. If your question is about Von Neumann machines, then we need to seperate between (logic) byte addressing and physical interface - for example modern x86 CPUs have a physical interface of 8 or more bytes wide, while on a locical level they operate bytewise. and so on. There's no real answer to that.
– Raffzahn
4 hours ago
1
No modern general purpose computer is "byte oriented" (that's a bit of a nebulous term). For example, in the x86_64 architecture, data comes in 8 byte chunks (with each byte being 8 bits). Maybe you should focus on machines that are byte addressable.
– JeremyP
3 hours ago
 |Â
show 8 more comments
1
There are processors today which have "weirdo" sizes, for example PIC16 has 14-bit program memory.
– Wilson
4 hours ago
1
Oh and at least some modern-day Elbrus chips also have a BESM-6 compatibility mode, which is a 48-bit computer which does not have bytes.
– Wilson
4 hours ago
3
A lot of modern DSPs (e.g. TI TMS 32000 series) use "bytes" that have 16 bits: processors.wiki.ti.com/index.php/…
– tofro
4 hours ago
2
Many Harvard type CPUs use different size for programm and data memory, so PICs can be had in 12, 14 and 16 bit program word size. If your question is about Von Neumann machines, then we need to seperate between (logic) byte addressing and physical interface - for example modern x86 CPUs have a physical interface of 8 or more bytes wide, while on a locical level they operate bytewise. and so on. There's no real answer to that.
– Raffzahn
4 hours ago
1
No modern general purpose computer is "byte oriented" (that's a bit of a nebulous term). For example, in the x86_64 architecture, data comes in 8 byte chunks (with each byte being 8 bits). Maybe you should focus on machines that are byte addressable.
– JeremyP
3 hours ago
1
1
There are processors today which have "weirdo" sizes, for example PIC16 has 14-bit program memory.
– Wilson
4 hours ago
There are processors today which have "weirdo" sizes, for example PIC16 has 14-bit program memory.
– Wilson
4 hours ago
1
1
Oh and at least some modern-day Elbrus chips also have a BESM-6 compatibility mode, which is a 48-bit computer which does not have bytes.
– Wilson
4 hours ago
Oh and at least some modern-day Elbrus chips also have a BESM-6 compatibility mode, which is a 48-bit computer which does not have bytes.
– Wilson
4 hours ago
3
3
A lot of modern DSPs (e.g. TI TMS 32000 series) use "bytes" that have 16 bits: processors.wiki.ti.com/index.php/…
– tofro
4 hours ago
A lot of modern DSPs (e.g. TI TMS 32000 series) use "bytes" that have 16 bits: processors.wiki.ti.com/index.php/…
– tofro
4 hours ago
2
2
Many Harvard type CPUs use different size for programm and data memory, so PICs can be had in 12, 14 and 16 bit program word size. If your question is about Von Neumann machines, then we need to seperate between (logic) byte addressing and physical interface - for example modern x86 CPUs have a physical interface of 8 or more bytes wide, while on a locical level they operate bytewise. and so on. There's no real answer to that.
– Raffzahn
4 hours ago
Many Harvard type CPUs use different size for programm and data memory, so PICs can be had in 12, 14 and 16 bit program word size. If your question is about Von Neumann machines, then we need to seperate between (logic) byte addressing and physical interface - for example modern x86 CPUs have a physical interface of 8 or more bytes wide, while on a locical level they operate bytewise. and so on. There's no real answer to that.
– Raffzahn
4 hours ago
1
1
No modern general purpose computer is "byte oriented" (that's a bit of a nebulous term). For example, in the x86_64 architecture, data comes in 8 byte chunks (with each byte being 8 bits). Maybe you should focus on machines that are byte addressable.
– JeremyP
3 hours ago
No modern general purpose computer is "byte oriented" (that's a bit of a nebulous term). For example, in the x86_64 architecture, data comes in 8 byte chunks (with each byte being 8 bits). Maybe you should focus on machines that are byte addressable.
– JeremyP
3 hours ago
 |Â
show 8 more comments
1 Answer
1
active
oldest
votes
up vote
3
down vote
Such a queston is a bit difficult, or rather impossible, to answer. While it is true that most mainstream computers today use units of 8 bits for bytes and and, at least Latin, characters, there always have been and still are exceptions. So, the answer to your "the last one" question probably is "there is none".
There are a number of wide-spread embedded MCUs with Harvard architecture that use 12-, 14- (PIC) or 16-bit wide (AVR) program memory and disallow 8-bit access to this memory. A "byte" in program memory for those MCUs thus has the above width.
The same thing applies for a lot of DSPs - They have byte widths of typically between 16 and 24 bits and very rarely allow (8-bit-)byte extraction from this memory. Typical examples would be the TMS 32000 (TI) or DSP56000 (Motorola/Freescale/NXP)
It is disputable whether systems based on either of these MCUs/DSPs should be considered "computers", but in my book they have to.
In a less strict sense, even relatively modern CPUs like MIPS could be considered to use, in this specific case, 32-bit "bytes". While the MIPS architecture has the concept of "8-bit bytes" in internal registers, MIPS CPUs technically cannot do less that 32-bit data transfers from and to memory. A similar restriction applies to the address registers in the Motorola 68k.
What seems to have evolved as a kind of standard, though, is that register and data bus width on most of today's CPUs typically is a multiple of eight.
answered 1 hour ago
tofro
12.7k32672
Thanks. This might be as good as I can expect but I'll wait a couple of more days in case there are some other interesting responses.
– badjohn
1 hour ago
The 68000 had 16 bit to/from memory restriction,60020 and up had 32 bit. 68008 had 8 bit databus
– UncleBod
1 hour ago
1
@UncleBod That is not what I meant. What you're referring to is the data bus width (and the 68020 had that configurable, BTW), but you can still load an 8-bit byte into any data register on any 68k CPU. But not into an address register, as these are limited to 16 bit transfers.
– tofro
1 hour ago
@torfo that makes it clearer.
– UncleBod
1 hour ago
PIC is definitely uses 8-bit bytes as data, but not as program memory. It does allow storing 8-bit data in its program memory, while not as straighforward (specifically, 14-bit-program-word-sized PICs store single 8-bit byte per program word). AVR, while having 16-bit program bus and 16- or 32-bit instructions, is still able to read its own program memory as bytes (thus, it stores 2 bytes per single word of program memory).
– lvd
19 mins ago
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
Such a queston is a bit difficult, or rather impossible, to answer. While it is true that most mainstream computers today use units of 8 bits for bytes and and, at least Latin, characters, there always have been and still are exceptions. So, the answer to your "the last one" question probably is "there is none".
There are a number of wide-spread embedded MCUs with Harvard architecture that use 12-, 14- (PIC) or 16-bit wide (AVR) program memory and disallow 8-bit access to this memory. A "byte" in program memory for those MCUs thus has the above width.
The same thing applies for a lot of DSPs - They have byte widths of typically between 16 and 24 bits and very rarely allow (8-bit-)byte extraction from this memory. Typical examples would be the TMS 32000 (TI) or DSP56000 (Motorola/Freescale/NXP)
It is disputable whether systems based on either of these MCUs/DSPs should be considered "computers", but in my book they have to.
In a less strict sense, even relatively modern CPUs like MIPS could be considered to use, in this specific case, 32-bit "bytes". While the MIPS architecture has the concept of "8-bit bytes" in internal registers, MIPS CPUs technically cannot do less that 32-bit data transfers from and to memory. A similar restriction applies to the address registers in the Motorola 68k.
What seems to have evolved as a kind of standard, though, is that register and data bus width on most of today's CPUs typically is a multiple of eight.
answered 1 hour ago
tofro
12.7k32672
Thanks. This might be as good as I can expect but I'll wait a couple of more days in case there are some other interesting responses.
– badjohn
1 hour ago
The 68000 had 16 bit to/from memory restriction,60020 and up had 32 bit. 68008 had 8 bit databus
– UncleBod
1 hour ago
1
@UncleBod That is not what I meant. What you're referring to is the data bus width (and the 68020 had that configurable, BTW), but you can still load an 8-bit byte into any data register on any 68k CPU. But not into an address register, as these are limited to 16 bit transfers.
– tofro
1 hour ago
@torfo that makes it clearer.
– UncleBod
1 hour ago
PIC is definitely uses 8-bit bytes as data, but not as program memory. It does allow storing 8-bit data in its program memory, while not as straighforward (specifically, 14-bit-program-word-sized PICs store single 8-bit byte per program word). AVR, while having 16-bit program bus and 16- or 32-bit instructions, is still able to read its own program memory as bytes (thus, it stores 2 bytes per single word of program memory).
– lvd
19 mins ago
add a comment |Â
up vote
3
down vote
Such a queston is a bit difficult, or rather impossible, to answer. While it is true that most mainstream computers today use units of 8 bits for bytes and and, at least Latin, characters, there always have been and still are exceptions. So, the answer to your "the last one" question probably is "there is none".
There are a number of wide-spread embedded MCUs with Harvard architecture that use 12-, 14- (PIC) or 16-bit wide (AVR) program memory and disallow 8-bit access to this memory. A "byte" in program memory for those MCUs thus has the above width.
The same thing applies for a lot of DSPs - They have byte widths of typically between 16 and 24 bits and very rarely allow (8-bit-)byte extraction from this memory. Typical examples would be the TMS 32000 (TI) or DSP56000 (Motorola/Freescale/NXP)
It is disputable whether systems based on either of these MCUs/DSPs should be considered "computers", but in my book they have to.
In a less strict sense, even relatively modern CPUs like MIPS could be considered to use, in this specific case, 32-bit "bytes". While the MIPS architecture has the concept of "8-bit bytes" in internal registers, MIPS CPUs technically cannot do less that 32-bit data transfers from and to memory. A similar restriction applies to the address registers in the Motorola 68k.
What seems to have evolved as a kind of standard, though, is that register and data bus width on most of today's CPUs typically is a multiple of eight.
answered 1 hour ago
tofro
12.7k32672
Thanks. This might be as good as I can expect but I'll wait a couple of more days in case there are some other interesting responses.
– badjohn
1 hour ago
The 68000 had 16 bit to/from memory restriction,60020 and up had 32 bit. 68008 had 8 bit databus
– UncleBod
1 hour ago
1
@UncleBod That is not what I meant. What you're referring to is the data bus width (and the 68020 had that configurable, BTW), but you can still load an 8-bit byte into any data register on any 68k CPU. But not into an address register, as these are limited to 16 bit transfers.
– tofro
1 hour ago
@torfo that makes it clearer.
– UncleBod
1 hour ago
PIC is definitely uses 8-bit bytes as data, but not as program memory. It does allow storing 8-bit data in its program memory, while not as straighforward (specifically, 14-bit-program-word-sized PICs store single 8-bit byte per program word). AVR, while having 16-bit program bus and 16- or 32-bit instructions, is still able to read its own program memory as bytes (thus, it stores 2 bytes per single word of program memory).
– lvd
19 mins ago
add a comment |Â
up vote
3
down vote
up vote
3
down vote
Such a queston is a bit difficult, or rather impossible, to answer. While it is true that most mainstream computers today use units of 8 bits for bytes and and, at least Latin, characters, there always have been and still are exceptions. So, the answer to your "the last one" question probably is "there is none".
There are a number of wide-spread embedded MCUs with Harvard architecture that use 12-, 14- (PIC) or 16-bit wide (AVR) program memory and disallow 8-bit access to this memory. A "byte" in program memory for those MCUs thus has the above width.
The same thing applies for a lot of DSPs - They have byte widths of typically between 16 and 24 bits and very rarely allow (8-bit-)byte extraction from this memory. Typical examples would be the TMS 32000 (TI) or DSP56000 (Motorola/Freescale/NXP)
It is disputable whether systems based on either of these MCUs/DSPs should be considered "computers", but in my book they have to.
In a less strict sense, even relatively modern CPUs like MIPS could be considered to use, in this specific case, 32-bit "bytes". While the MIPS architecture has the concept of "8-bit bytes" in internal registers, MIPS CPUs technically cannot do less that 32-bit data transfers from and to memory. A similar restriction applies to the address registers in the Motorola 68k.
What seems to have evolved as a kind of standard, though, is that register and data bus width on most of today's CPUs typically is a multiple of eight.
answered 1 hour ago
tofro
12.7k32672
Such a queston is a bit difficult, or rather impossible, to answer. While it is true that most mainstream computers today use units of 8 bits for bytes and and, at least Latin, characters, there always have been and still are exceptions. So, the answer to your "the last one" question probably is "there is none".
There are a number of wide-spread embedded MCUs with Harvard architecture that use 12-, 14- (PIC) or 16-bit wide (AVR) program memory and disallow 8-bit access to this memory. A "byte" in program memory for those MCUs thus has the above width.
The same thing applies for a lot of DSPs - They have byte widths of typically between 16 and 24 bits and very rarely allow (8-bit-)byte extraction from this memory. Typical examples would be the TMS 32000 (TI) or DSP56000 (Motorola/Freescale/NXP)
It is disputable whether systems based on either of these MCUs/DSPs should be considered "computers", but in my book they have to.
In a less strict sense, even relatively modern CPUs like MIPS could be considered to use, in this specific case, 32-bit "bytes". While the MIPS architecture has the concept of "8-bit bytes" in internal registers, MIPS CPUs technically cannot do less that 32-bit data transfers from and to memory. A similar restriction applies to the address registers in the Motorola 68k.
What seems to have evolved as a kind of standard, though, is that register and data bus width on most of today's CPUs typically is a multiple of eight.
answered 1 hour ago
tofro
12.7k32672
edited 1 hour ago
answered 1 hour ago
tofro
12.7k32672
answered 1 hour ago
tofro
12.7k32672
answered 1 hour ago
tofro
12.7k32672
12.7k32672
Thanks. This might be as good as I can expect but I'll wait a couple of more days in case there are some other interesting responses.
– badjohn
1 hour ago
The 68000 had 16 bit to/from memory restriction,60020 and up had 32 bit. 68008 had 8 bit databus
– UncleBod
1 hour ago
1
@UncleBod That is not what I meant. What you're referring to is the data bus width (and the 68020 had that configurable, BTW), but you can still load an 8-bit byte into any data register on any 68k CPU. But not into an address register, as these are limited to 16 bit transfers.
– tofro
1 hour ago
@torfo that makes it clearer.
– UncleBod
1 hour ago
PIC is definitely uses 8-bit bytes as data, but not as program memory. It does allow storing 8-bit data in its program memory, while not as straighforward (specifically, 14-bit-program-word-sized PICs store single 8-bit byte per program word). AVR, while having 16-bit program bus and 16- or 32-bit instructions, is still able to read its own program memory as bytes (thus, it stores 2 bytes per single word of program memory).
– lvd
19 mins ago
add a comment |Â
Thanks. This might be as good as I can expect but I'll wait a couple of more days in case there are some other interesting responses.
– badjohn
1 hour ago
The 68000 had 16 bit to/from memory restriction,60020 and up had 32 bit. 68008 had 8 bit databus
– UncleBod
1 hour ago
1
@UncleBod That is not what I meant. What you're referring to is the data bus width (and the 68020 had that configurable, BTW), but you can still load an 8-bit byte into any data register on any 68k CPU. But not into an address register, as these are limited to 16 bit transfers.
– tofro
1 hour ago
@torfo that makes it clearer.
– UncleBod
1 hour ago
PIC is definitely uses 8-bit bytes as data, but not as program memory. It does allow storing 8-bit data in its program memory, while not as straighforward (specifically, 14-bit-program-word-sized PICs store single 8-bit byte per program word). AVR, while having 16-bit program bus and 16- or 32-bit instructions, is still able to read its own program memory as bytes (thus, it stores 2 bytes per single word of program memory).
– lvd
19 mins ago
Thanks. This might be as good as I can expect but I'll wait a couple of more days in case there are some other interesting responses.
– badjohn
1 hour ago
Thanks. This might be as good as I can expect but I'll wait a couple of more days in case there are some other interesting responses.
– badjohn
1 hour ago
The 68000 had 16 bit to/from memory restriction,60020 and up had 32 bit. 68008 had 8 bit databus
– UncleBod
1 hour ago
The 68000 had 16 bit to/from memory restriction,60020 and up had 32 bit. 68008 had 8 bit databus
– UncleBod
1 hour ago
1
1
@UncleBod That is not what I meant. What you're referring to is the data bus width (and the 68020 had that configurable, BTW), but you can still load an 8-bit byte into any data register on any 68k CPU. But not into an address register, as these are limited to 16 bit transfers.
– tofro
1 hour ago
@UncleBod That is not what I meant. What you're referring to is the data bus width (and the 68020 had that configurable, BTW), but you can still load an 8-bit byte into any data register on any 68k CPU. But not into an address register, as these are limited to 16 bit transfers.
– tofro
1 hour ago
@torfo that makes it clearer.
– UncleBod
1 hour ago
@torfo that makes it clearer.
– UncleBod
1 hour ago
PIC is definitely uses 8-bit bytes as data, but not as program memory. It does allow storing 8-bit data in its program memory, while not as straighforward (specifically, 14-bit-program-word-sized PICs store single 8-bit byte per program word). AVR, while having 16-bit program bus and 16- or 32-bit instructions, is still able to read its own program memory as bytes (thus, it stores 2 bytes per single word of program memory).
– lvd
19 mins ago
PIC is definitely uses 8-bit bytes as data, but not as program memory. It does allow storing 8-bit data in its program memory, while not as straighforward (specifically, 14-bit-program-word-sized PICs store single 8-bit byte per program word). AVR, while having 16-bit program bus and 16- or 32-bit instructions, is still able to read its own program memory as bytes (thus, it stores 2 bytes per single word of program memory).
– lvd
19 mins ago
add a comment |Â
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1
There are processors today which have "weirdo" sizes, for example PIC16 has 14-bit program memory.
– Wilson
4 hours ago
1
Oh and at least some modern-day Elbrus chips also have a BESM-6 compatibility mode, which is a 48-bit computer which does not have bytes.
– Wilson
4 hours ago
3
A lot of modern DSPs (e.g. TI TMS 32000 series) use "bytes" that have 16 bits: processors.wiki.ti.com/index.php/…
– tofro
4 hours ago
2
Many Harvard type CPUs use different size for programm and data memory, so PICs can be had in 12, 14 and 16 bit program word size. If your question is about Von Neumann machines, then we need to seperate between (logic) byte addressing and physical interface - for example modern x86 CPUs have a physical interface of 8 or more bytes wide, while on a locical level they operate bytewise. and so on. There's no real answer to that.
– Raffzahn
4 hours ago
1
No modern general purpose computer is "byte oriented" (that's a bit of a nebulous term). For example, in the x86_64 architecture, data comes in 8 byte chunks (with each byte being 8 bits). Maybe you should focus on machines that are byte addressable.
– JeremyP
3 hours ago