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Last computer not to 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
edited 16 mins ago
Tommy
12.7k13264
asked 6 hours ago
badjohn
4939
 |Â
show 9 more comments
up vote
4
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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
edited 16 mins ago
Tommy
12.7k13264
asked 6 hours ago
badjohn
4939
1
There are processors today which have "weirdo" sizes, for example PIC16 has 14-bit program memory.
– Wilson
6 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
6 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
5 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
5 hours ago
2
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
4 hours ago
 |Â
show 9 more comments
up vote
4
down vote
favorite
up vote
4
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
edited 16 mins ago
Tommy
12.7k13264
asked 6 hours ago
badjohn
4939
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
edited 16 mins ago
Tommy
12.7k13264
asked 6 hours ago
badjohn
4939
edited 16 mins ago
Tommy
12.7k13264
asked 6 hours ago
badjohn
4939
edited 16 mins ago
Tommy
12.7k13264
edited 16 mins ago
Tommy
12.7k13264
edited 16 mins ago
Tommy
12.7k13264
12.7k13264
asked 6 hours ago
badjohn
4939
asked 6 hours ago
badjohn
4939
asked 6 hours ago
badjohn
4939
4939
1
There are processors today which have "weirdo" sizes, for example PIC16 has 14-bit program memory.
– Wilson
6 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
6 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
5 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
5 hours ago
2
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
4 hours ago
 |Â
show 9 more comments
1
There are processors today which have "weirdo" sizes, for example PIC16 has 14-bit program memory.
– Wilson
6 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
6 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
5 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
5 hours ago
2
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
4 hours ago
1
1
There are processors today which have "weirdo" sizes, for example PIC16 has 14-bit program memory.
– Wilson
6 hours ago
There are processors today which have "weirdo" sizes, for example PIC16 has 14-bit program memory.
– Wilson
6 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
6 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
6 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
5 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
5 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
5 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
5 hours ago
2
2
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
4 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
4 hours ago
 |Â
show 9 more comments
3 Answers
3
active
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up vote
5
down vote
Such a question 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 widespread 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 than 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.
edited 31 mins ago
manassehkatz
1,311111
answered 3 hours 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
3 hours ago
The 68000 had 16 bit to/from memory restriction,60020 and up had 32 bit. 68008 had 8 bit databus
– UncleBod
2 hours 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
2 hours ago
@torfo that makes it clearer.
– UncleBod
2 hours 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
1 hour ago
 |Â
show 1 more comment
up vote
2
down vote
Univac 36-bit
The Univac 1100/2200 series used a 36-bit word. Many models had functions to work with a word as 4 9-bit "bytes" - using the term byte but fitting your definition of a non-byte computer based on not using 8-bit bytes. I had a few courses on an 1100/80 at the University of Maryland in the early 80s.
The 2200 series appears to have carried on the 36-bit architecture, which in theory means up to at least 1997 according to the Wikipedia article. But the ClearPath series, while compatible with the 1100/2200, also included Xeon which would, by definition, mean it included plenty of 8-bit byte oriented instructions.
answered 1 hour ago
manassehkatz
1,311111
add a comment |Â
up vote
2
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In the early 1990's CDC sold a line of Cyber 180 mainframes. These machines were descendants of the CDC 6600 and supported that machine's 60-bit word size and 6-bit characters. Notably, one of the innovations of the 180 over the 170 is that the 180 added support for 64-bit words and 8-bit characters, and could run software written for both modes simultaneously. So this is probably towards the tail end of sub-8-bit character oriented computing, at least at commercial data processing scale.
answered 13 mins ago
mschaef
1,871513
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
Such a question 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 widespread 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 than 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.
edited 31 mins ago
manassehkatz
1,311111
answered 3 hours 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
3 hours ago
The 68000 had 16 bit to/from memory restriction,60020 and up had 32 bit. 68008 had 8 bit databus
– UncleBod
2 hours 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
2 hours ago
@torfo that makes it clearer.
– UncleBod
2 hours 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
1 hour ago
 |Â
show 1 more comment
up vote
5
down vote
Such a question 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 widespread 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 than 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.
edited 31 mins ago
manassehkatz
1,311111
answered 3 hours 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
3 hours ago
The 68000 had 16 bit to/from memory restriction,60020 and up had 32 bit. 68008 had 8 bit databus
– UncleBod
2 hours 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
2 hours ago
@torfo that makes it clearer.
– UncleBod
2 hours 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
1 hour ago
 |Â
show 1 more comment
up vote
5
down vote
up vote
5
down vote
Such a question 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 widespread 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 than 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.
edited 31 mins ago
manassehkatz
1,311111
answered 3 hours ago
tofro
12.7k32672
Such a question 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 widespread 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 than 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.
edited 31 mins ago
manassehkatz
1,311111
answered 3 hours ago
tofro
12.7k32672
edited 31 mins ago
manassehkatz
1,311111
edited 31 mins ago
manassehkatz
1,311111
edited 31 mins ago
manassehkatz
1,311111
1,311111
answered 3 hours ago
tofro
12.7k32672
answered 3 hours ago
tofro
12.7k32672
answered 3 hours 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
3 hours ago
The 68000 had 16 bit to/from memory restriction,60020 and up had 32 bit. 68008 had 8 bit databus
– UncleBod
2 hours 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
2 hours ago
@torfo that makes it clearer.
– UncleBod
2 hours 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
1 hour ago
 |Â
show 1 more comment
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
3 hours ago
The 68000 had 16 bit to/from memory restriction,60020 and up had 32 bit. 68008 had 8 bit databus
– UncleBod
2 hours 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
2 hours ago
@torfo that makes it clearer.
– UncleBod
2 hours 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
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
3 hours 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
3 hours ago
The 68000 had 16 bit to/from memory restriction,60020 and up had 32 bit. 68008 had 8 bit databus
– UncleBod
2 hours ago
The 68000 had 16 bit to/from memory restriction,60020 and up had 32 bit. 68008 had 8 bit databus
– UncleBod
2 hours 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
2 hours 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
2 hours ago
@torfo that makes it clearer.
– UncleBod
2 hours ago
@torfo that makes it clearer.
– UncleBod
2 hours 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
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
1 hour ago
 |Â
show 1 more comment
up vote
2
down vote
Univac 36-bit
The Univac 1100/2200 series used a 36-bit word. Many models had functions to work with a word as 4 9-bit "bytes" - using the term byte but fitting your definition of a non-byte computer based on not using 8-bit bytes. I had a few courses on an 1100/80 at the University of Maryland in the early 80s.
The 2200 series appears to have carried on the 36-bit architecture, which in theory means up to at least 1997 according to the Wikipedia article. But the ClearPath series, while compatible with the 1100/2200, also included Xeon which would, by definition, mean it included plenty of 8-bit byte oriented instructions.
answered 1 hour ago
manassehkatz
1,311111
add a comment |Â
up vote
2
down vote
Univac 36-bit
The Univac 1100/2200 series used a 36-bit word. Many models had functions to work with a word as 4 9-bit "bytes" - using the term byte but fitting your definition of a non-byte computer based on not using 8-bit bytes. I had a few courses on an 1100/80 at the University of Maryland in the early 80s.
The 2200 series appears to have carried on the 36-bit architecture, which in theory means up to at least 1997 according to the Wikipedia article. But the ClearPath series, while compatible with the 1100/2200, also included Xeon which would, by definition, mean it included plenty of 8-bit byte oriented instructions.
answered 1 hour ago
manassehkatz
1,311111
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up vote
2
down vote
up vote
2
down vote
Univac 36-bit
The Univac 1100/2200 series used a 36-bit word. Many models had functions to work with a word as 4 9-bit "bytes" - using the term byte but fitting your definition of a non-byte computer based on not using 8-bit bytes. I had a few courses on an 1100/80 at the University of Maryland in the early 80s.
The 2200 series appears to have carried on the 36-bit architecture, which in theory means up to at least 1997 according to the Wikipedia article. But the ClearPath series, while compatible with the 1100/2200, also included Xeon which would, by definition, mean it included plenty of 8-bit byte oriented instructions.
answered 1 hour ago
manassehkatz
1,311111
Univac 36-bit
The Univac 1100/2200 series used a 36-bit word. Many models had functions to work with a word as 4 9-bit "bytes" - using the term byte but fitting your definition of a non-byte computer based on not using 8-bit bytes. I had a few courses on an 1100/80 at the University of Maryland in the early 80s.
The 2200 series appears to have carried on the 36-bit architecture, which in theory means up to at least 1997 according to the Wikipedia article. But the ClearPath series, while compatible with the 1100/2200, also included Xeon which would, by definition, mean it included plenty of 8-bit byte oriented instructions.
answered 1 hour ago
manassehkatz
1,311111
answered 1 hour ago
manassehkatz
1,311111
answered 1 hour ago
manassehkatz
1,311111
answered 1 hour ago
manassehkatz
1,311111
1,311111
add a comment |Â
add a comment |Â
up vote
2
down vote
In the early 1990's CDC sold a line of Cyber 180 mainframes. These machines were descendants of the CDC 6600 and supported that machine's 60-bit word size and 6-bit characters. Notably, one of the innovations of the 180 over the 170 is that the 180 added support for 64-bit words and 8-bit characters, and could run software written for both modes simultaneously. So this is probably towards the tail end of sub-8-bit character oriented computing, at least at commercial data processing scale.
answered 13 mins ago
mschaef
1,871513
add a comment |Â
up vote
2
down vote
In the early 1990's CDC sold a line of Cyber 180 mainframes. These machines were descendants of the CDC 6600 and supported that machine's 60-bit word size and 6-bit characters. Notably, one of the innovations of the 180 over the 170 is that the 180 added support for 64-bit words and 8-bit characters, and could run software written for both modes simultaneously. So this is probably towards the tail end of sub-8-bit character oriented computing, at least at commercial data processing scale.
answered 13 mins ago
mschaef
1,871513
add a comment |Â
up vote
2
down vote
up vote
2
down vote
In the early 1990's CDC sold a line of Cyber 180 mainframes. These machines were descendants of the CDC 6600 and supported that machine's 60-bit word size and 6-bit characters. Notably, one of the innovations of the 180 over the 170 is that the 180 added support for 64-bit words and 8-bit characters, and could run software written for both modes simultaneously. So this is probably towards the tail end of sub-8-bit character oriented computing, at least at commercial data processing scale.
answered 13 mins ago
mschaef
1,871513
In the early 1990's CDC sold a line of Cyber 180 mainframes. These machines were descendants of the CDC 6600 and supported that machine's 60-bit word size and 6-bit characters. Notably, one of the innovations of the 180 over the 170 is that the 180 added support for 64-bit words and 8-bit characters, and could run software written for both modes simultaneously. So this is probably towards the tail end of sub-8-bit character oriented computing, at least at commercial data processing scale.
answered 13 mins ago
mschaef
1,871513
answered 13 mins ago
mschaef
1,871513
answered 13 mins ago
mschaef
1,871513
answered 13 mins ago
mschaef
1,871513
1,871513
add a comment |Â
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
6 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
6 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
5 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
5 hours ago
2
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
4 hours ago