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Ugo
25 days ago
13

On a typical microprocessor, a distinct I/O address is used to refer to the I/O data registers and a distinct address for the co

ntrol and status registers in an I/O controller for a given device. Such registers are referred to as ports. In the Intel 8088, two I/O instruction formats are used. In one format, the 8-bit opcode specifies an I/O operation; this is followed by an 8-bit port address. Other I/O opcodes imply that the port address is in the 16-bit DX register. How many ports can the 8088 address in each I/O addressing mode?
Computers and Technology
1 answer:
maria [879]25 days ago
5 0

Answer:

First I/O instruction format -> 256 ports

Second I/O instruction format -> 65536 ports

Explanation:

  • The initial instruction format can address 256 ports, computed from eight bits for port addressing as 2^8 = 256 ports.
  • The second instruction format allows for 65536 ports, which can be calculated using sixteen bits for addressing as 2^16 = 65536 ports.

Changing the opcode facilitates the selection between the first and second instruction format, allowing one input or output operation at a given moment.

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To mitigate or minimize collisions in a network, modern wired networks utilize network switches where each device is attached to its individual port on the switch. This configuration turns the switch into a collision domain for half duplex connections, while in the case of full duplex links, the risk of collisions is entirely removed.

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