Question

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 control 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?

Answer 1

Solution

Before going to solution

It is mentioned in the question 8088

therefore in 8088 machines there are 2 types of I/O instruction formats are available

1st Instruction format

• No.of Bits for Input/Output operation-8 bits
• No.of bits for port addressing-8 bit

2nd instruction format

• port address is 16bit DX register

Lets see one by one

Number of ports for the 1st instruction format

No.of Bits used for the purpose of port addressing-8bits

8 bits means the no.of ports addressed is

=28

=256 ports

it will allow 128 devices to be addressed the reason is that opcode either allow input or output function at a time but not both thats why it is possible to reuse this addresses

Therefore

256 input port addresses

256 output port addresses

so

No.of ports the 1st instruction format can addres up to 256 ports

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Number of ports for 2nd instruction format

No.of bits used for port addressing is 16 bits

with 8 bits no.of ports can be addressed

216=26*210

=64 K

(210 means 1K)

so

No.of ports 2nd instruction format can address upt to 64K ports

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all the best