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Why do the i8080 I/O instructions take a byte-sized operand to determine the port?
Why did Pokémon Red have so many overflow bugs?DIY Project using 1970s 8-bit CPU, is it possible?What is the relative code density of 8-bit microprocessors?Why was the 8080's JumP immediate instruction placed where it was?Intel 8080 - Behaviour of the carry bit when comparing a value with 0The start of x86: Intel 8080 vs Intel 8086?Intel 8080 and Altair 8800. 256 I/0 ports, but only 7 free RST (interrupt subroutine) - how it works?What is the difference between an 8080 and an 8051?Comparing Signed Numbers on Z80 (8080) in AssemblyWhere can I find and download emulator of Altair 8800 with Intel 8080 processor for Windows?
I thought the i8080 had 8 16-bit IN ports and 8 16-bit OUT ports. Doesn't that mean the instructions
IN d8 ; only index 0-7?
OUT d8
can only take 8 possible values? What would it be indexing otherwise?
8080
asked 6 hours ago
David TranDavid Tran
82
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David Tran is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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I thought the i8080 had 8 16-bit IN ports and 8 16-bit OUT ports. Doesn't that mean the instructions
IN d8 ; only index 0-7?
OUT d8
can only take 8 possible values? What would it be indexing otherwise?
8080
asked 6 hours ago
David TranDavid Tran
82
New contributor
David Tran is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
I thought the i8080 had 8 16-bit IN ports and 8 16-bit OUT ports. Doesn't that mean the instructions
IN d8 ; only index 0-7?
OUT d8
can only take 8 possible values? What would it be indexing otherwise?
8080
asked 6 hours ago
David TranDavid Tran
82
New contributor
David Tran is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I thought the i8080 had 8 16-bit IN ports and 8 16-bit OUT ports. Doesn't that mean the instructions
IN d8 ; only index 0-7?
OUT d8
can only take 8 possible values? What would it be indexing otherwise?
8080
8080
asked 6 hours ago
David TranDavid Tran
82
New contributor
David Tran is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 6 hours ago
David TranDavid Tran
82
New contributor
David Tran is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 6 hours ago
David TranDavid Tran
82
New contributor
David Tran is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 6 hours ago
David TranDavid Tran
82
asked 6 hours ago
David TranDavid Tran
82
82
New contributor
David Tran is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
David Tran is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
2 Answers
2
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oldest
votes
I thought the i8080 had 8 16-bit IN ports and 8 16-bit OUT ports.
The 8080 does not have any I/O Ports. It's a microprocessor, not a microcontroller.
(Maybe the system you're playing with does have these 8+8 ports, but they are always external to the CPU)
The 8080 features a 16 bit data/program address space and an 8 bit I/O i/o address space. Or in other words it can address 64KiB of memory and 256 I/O locations. Each of them wit 8 Bit.
answered 6 hours ago
RaffzahnRaffzahn
58.2k6142237
1
Which locations areinand which areout? Is it like 0-127 and 128-256 kinda thing?
– David Tran
6 hours ago
No. it's an address space. Not ports. What responds to each address, and if its read only, write only or read/write depends on the I/O unit assigned to that address. Keep in mind, the 8080 is a microprocessor, not a microcontroller there are no build in ports.
– Raffzahn
6 hours ago
Ahhh okay thank you. Makes sense now.
– David Tran
6 hours ago
add a comment |
The question of whether something is "memory" or "I/O" depends upon how it responds to various control signals. There is nothing that would prevent anyone from wiring an I/O device so that it would respond to a range of "memory" addresses, and for some kinds of I/O device that could be more useful than wiring it to the "I/O" read/write signals. On the other hand, an I/O device which is wired to behave as a memory device will respond to an instruction that writes to HL whenever HL holds its address, without regard for whether HL was supposed to hold that address. An I/O device that response to I/O address 0x57 by contrast will only respond to an "out 57h" instruction, and the likelihood of the processor encountering that byte sequence by chance is smaller than the likelihood that HL might end up with a bogus address.
While it would have been possible to make the OUT nn instruction take a two-byte address operand, that would have made the opcode bigger and slower while offering little benefit for most applications, especially given that the applications which would need more than 256 bytes of I/O space would also need to access I/O devices using register-based addresses and should thus likely be wired as "memory" devices.
answered 2 hours ago
supercatsupercat
8,7701144
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
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votes
I thought the i8080 had 8 16-bit IN ports and 8 16-bit OUT ports.
The 8080 does not have any I/O Ports. It's a microprocessor, not a microcontroller.
(Maybe the system you're playing with does have these 8+8 ports, but they are always external to the CPU)
The 8080 features a 16 bit data/program address space and an 8 bit I/O i/o address space. Or in other words it can address 64KiB of memory and 256 I/O locations. Each of them wit 8 Bit.
answered 6 hours ago
RaffzahnRaffzahn
58.2k6142237
1
Which locations areinand which areout? Is it like 0-127 and 128-256 kinda thing?
– David Tran
6 hours ago
No. it's an address space. Not ports. What responds to each address, and if its read only, write only or read/write depends on the I/O unit assigned to that address. Keep in mind, the 8080 is a microprocessor, not a microcontroller there are no build in ports.
– Raffzahn
6 hours ago
Ahhh okay thank you. Makes sense now.
– David Tran
6 hours ago
add a comment |
I thought the i8080 had 8 16-bit IN ports and 8 16-bit OUT ports.
The 8080 does not have any I/O Ports. It's a microprocessor, not a microcontroller.
(Maybe the system you're playing with does have these 8+8 ports, but they are always external to the CPU)
The 8080 features a 16 bit data/program address space and an 8 bit I/O i/o address space. Or in other words it can address 64KiB of memory and 256 I/O locations. Each of them wit 8 Bit.
answered 6 hours ago
RaffzahnRaffzahn
58.2k6142237
1
Which locations areinand which areout? Is it like 0-127 and 128-256 kinda thing?
– David Tran
6 hours ago
No. it's an address space. Not ports. What responds to each address, and if its read only, write only or read/write depends on the I/O unit assigned to that address. Keep in mind, the 8080 is a microprocessor, not a microcontroller there are no build in ports.
– Raffzahn
6 hours ago
Ahhh okay thank you. Makes sense now.
– David Tran
6 hours ago
add a comment |
I thought the i8080 had 8 16-bit IN ports and 8 16-bit OUT ports.
The 8080 does not have any I/O Ports. It's a microprocessor, not a microcontroller.
(Maybe the system you're playing with does have these 8+8 ports, but they are always external to the CPU)
The 8080 features a 16 bit data/program address space and an 8 bit I/O i/o address space. Or in other words it can address 64KiB of memory and 256 I/O locations. Each of them wit 8 Bit.
answered 6 hours ago
RaffzahnRaffzahn
58.2k6142237
I thought the i8080 had 8 16-bit IN ports and 8 16-bit OUT ports.
The 8080 does not have any I/O Ports. It's a microprocessor, not a microcontroller.
(Maybe the system you're playing with does have these 8+8 ports, but they are always external to the CPU)
The 8080 features a 16 bit data/program address space and an 8 bit I/O i/o address space. Or in other words it can address 64KiB of memory and 256 I/O locations. Each of them wit 8 Bit.
answered 6 hours ago
RaffzahnRaffzahn
58.2k6142237
edited 4 hours ago
answered 6 hours ago
RaffzahnRaffzahn
58.2k6142237
answered 6 hours ago
RaffzahnRaffzahn
58.2k6142237
answered 6 hours ago
RaffzahnRaffzahn
58.2k6142237
58.2k6142237
1
Which locations areinand which areout? Is it like 0-127 and 128-256 kinda thing?
– David Tran
6 hours ago
No. it's an address space. Not ports. What responds to each address, and if its read only, write only or read/write depends on the I/O unit assigned to that address. Keep in mind, the 8080 is a microprocessor, not a microcontroller there are no build in ports.
– Raffzahn
6 hours ago
Ahhh okay thank you. Makes sense now.
– David Tran
6 hours ago
add a comment |
1
Which locations areinand which areout? Is it like 0-127 and 128-256 kinda thing?
– David Tran
6 hours ago
No. it's an address space. Not ports. What responds to each address, and if its read only, write only or read/write depends on the I/O unit assigned to that address. Keep in mind, the 8080 is a microprocessor, not a microcontroller there are no build in ports.
– Raffzahn
6 hours ago
Ahhh okay thank you. Makes sense now.
– David Tran
6 hours ago
1
1
Which locations are
in and which are out? Is it like 0-127 and 128-256 kinda thing?– David Tran
6 hours ago
Which locations are
in and which are out? Is it like 0-127 and 128-256 kinda thing?– David Tran
6 hours ago
No. it's an address space. Not ports. What responds to each address, and if its read only, write only or read/write depends on the I/O unit assigned to that address. Keep in mind, the 8080 is a microprocessor, not a microcontroller there are no build in ports.
– Raffzahn
6 hours ago
No. it's an address space. Not ports. What responds to each address, and if its read only, write only or read/write depends on the I/O unit assigned to that address. Keep in mind, the 8080 is a microprocessor, not a microcontroller there are no build in ports.
– Raffzahn
6 hours ago
Ahhh okay thank you. Makes sense now.
– David Tran
6 hours ago
Ahhh okay thank you. Makes sense now.
– David Tran
6 hours ago
add a comment |
The question of whether something is "memory" or "I/O" depends upon how it responds to various control signals. There is nothing that would prevent anyone from wiring an I/O device so that it would respond to a range of "memory" addresses, and for some kinds of I/O device that could be more useful than wiring it to the "I/O" read/write signals. On the other hand, an I/O device which is wired to behave as a memory device will respond to an instruction that writes to HL whenever HL holds its address, without regard for whether HL was supposed to hold that address. An I/O device that response to I/O address 0x57 by contrast will only respond to an "out 57h" instruction, and the likelihood of the processor encountering that byte sequence by chance is smaller than the likelihood that HL might end up with a bogus address.
While it would have been possible to make the OUT nn instruction take a two-byte address operand, that would have made the opcode bigger and slower while offering little benefit for most applications, especially given that the applications which would need more than 256 bytes of I/O space would also need to access I/O devices using register-based addresses and should thus likely be wired as "memory" devices.
answered 2 hours ago
supercatsupercat
8,7701144
add a comment |
The question of whether something is "memory" or "I/O" depends upon how it responds to various control signals. There is nothing that would prevent anyone from wiring an I/O device so that it would respond to a range of "memory" addresses, and for some kinds of I/O device that could be more useful than wiring it to the "I/O" read/write signals. On the other hand, an I/O device which is wired to behave as a memory device will respond to an instruction that writes to HL whenever HL holds its address, without regard for whether HL was supposed to hold that address. An I/O device that response to I/O address 0x57 by contrast will only respond to an "out 57h" instruction, and the likelihood of the processor encountering that byte sequence by chance is smaller than the likelihood that HL might end up with a bogus address.
While it would have been possible to make the OUT nn instruction take a two-byte address operand, that would have made the opcode bigger and slower while offering little benefit for most applications, especially given that the applications which would need more than 256 bytes of I/O space would also need to access I/O devices using register-based addresses and should thus likely be wired as "memory" devices.
answered 2 hours ago
supercatsupercat
8,7701144
add a comment |
The question of whether something is "memory" or "I/O" depends upon how it responds to various control signals. There is nothing that would prevent anyone from wiring an I/O device so that it would respond to a range of "memory" addresses, and for some kinds of I/O device that could be more useful than wiring it to the "I/O" read/write signals. On the other hand, an I/O device which is wired to behave as a memory device will respond to an instruction that writes to HL whenever HL holds its address, without regard for whether HL was supposed to hold that address. An I/O device that response to I/O address 0x57 by contrast will only respond to an "out 57h" instruction, and the likelihood of the processor encountering that byte sequence by chance is smaller than the likelihood that HL might end up with a bogus address.
While it would have been possible to make the OUT nn instruction take a two-byte address operand, that would have made the opcode bigger and slower while offering little benefit for most applications, especially given that the applications which would need more than 256 bytes of I/O space would also need to access I/O devices using register-based addresses and should thus likely be wired as "memory" devices.
answered 2 hours ago
supercatsupercat
8,7701144
The question of whether something is "memory" or "I/O" depends upon how it responds to various control signals. There is nothing that would prevent anyone from wiring an I/O device so that it would respond to a range of "memory" addresses, and for some kinds of I/O device that could be more useful than wiring it to the "I/O" read/write signals. On the other hand, an I/O device which is wired to behave as a memory device will respond to an instruction that writes to HL whenever HL holds its address, without regard for whether HL was supposed to hold that address. An I/O device that response to I/O address 0x57 by contrast will only respond to an "out 57h" instruction, and the likelihood of the processor encountering that byte sequence by chance is smaller than the likelihood that HL might end up with a bogus address.
While it would have been possible to make the OUT nn instruction take a two-byte address operand, that would have made the opcode bigger and slower while offering little benefit for most applications, especially given that the applications which would need more than 256 bytes of I/O space would also need to access I/O devices using register-based addresses and should thus likely be wired as "memory" devices.
answered 2 hours ago
supercatsupercat
8,7701144
answered 2 hours ago
supercatsupercat
8,7701144
answered 2 hours ago
supercatsupercat
8,7701144
answered 2 hours ago
supercatsupercat
8,7701144
8,7701144
add a comment |
add a comment |
David Tran is a new contributor. Be nice, and check out our Code of Conduct.
David Tran is a new contributor. Be nice, and check out our Code of Conduct.
David Tran is a new contributor. Be nice, and check out our Code of Conduct.
David Tran is a new contributor. Be nice, and check out our Code of Conduct.
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