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Why do the i8080 I/O instructions take a byte-sized operand to determine the port?

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2

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

share|improve this question

asked 7 hours ago

David TranDavid Tran

132

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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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2

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

share|improve this question

asked 7 hours ago

David TranDavid Tran

132

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

share|improve this question

asked 7 hours ago

David TranDavid Tran

132

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.

IN d8 ; only index 0-7?

OUT d8 

share|improve this question

asked 7 hours ago

David TranDavid Tran

132

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.

share|improve this question

asked 7 hours ago

David TranDavid Tran

132

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 7 hours ago

David TranDavid Tran

132

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 7 hours ago

David TranDavid Tran

132

2 Answers
2

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5

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.

share|improve this answer

edited 5 hours ago

answered 7 hours ago

RaffzahnRaffzahn

58.2k6142237

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Which locations are in and which are out? Is it like 0-127 and 128-256 kinda thing?
  
  – David Tran
  7 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
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Ahhh okay thank you. Makes sense now.
  
  – David Tran
  7 hours ago
  

  
  
  
  

add a comment | 

1

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.

share|improve this answer

answered 3 hours ago

supercatsupercat

8,7701144

add a comment | 

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5

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.

share|improve this answer

edited 5 hours ago

answered 7 hours ago

RaffzahnRaffzahn

58.2k6142237

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Which locations are in and which are out? Is it like 0-127 and 128-256 kinda thing?
  
  – David Tran
  7 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
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Ahhh okay thank you. Makes sense now.
  
  – David Tran
  7 hours ago
  

  
  
  
  

add a comment | 

5

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.

share|improve this answer

edited 5 hours ago

answered 7 hours ago

RaffzahnRaffzahn

58.2k6142237

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  Which locations are in and which are out? Is it like 0-127 and 128-256 kinda thing?
  
  – David Tran
  7 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
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Ahhh okay thank you. Makes sense now.
  
  – David Tran
  7 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.

share|improve this answer

edited 5 hours ago

answered 7 hours ago

RaffzahnRaffzahn

58.2k6142237

share|improve this answer

edited 5 hours ago

answered 7 hours ago

RaffzahnRaffzahn

58.2k6142237

answered 7 hours ago

RaffzahnRaffzahn

58.2k6142237

answered 7 hours ago

RaffzahnRaffzahn

58.2k6142237

1

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.

share|improve this answer

answered 3 hours ago

supercatsupercat

8,7701144

add a comment | 

1

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.

share|improve this answer

answered 3 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.

share|improve this answer

answered 3 hours ago

supercatsupercat

8,7701144

share|improve this answer

answered 3 hours ago

supercatsupercat

8,7701144

answered 3 hours ago

supercatsupercat

8,7701144

answered 3 hours ago

supercatsupercat

8,7701144