Chapter 6, Problem 33E

Explanation of Solution

Consider the given instructions:

  0001 D0 00 48

  0004 F1 FC 16

Executing the first instruction:

Consider the first instruction “D0 00 48” stored in the address “0001”.

• Here, the hexadecimal value “D0” is an instruction and “00 48” is the operand.
• The binary equivalent of the instruction “D0” is “1101 0000”, which will be stored in the instruction specifier.
• In the instruction specifier, the first four bits refers the operation code, the fifth bit refers the register specifier, and the next three bits refers the addressing mode.
  • The first four bits “1101” – Load byte into the “A” register.
  • The fifth bit “0” – Indicates the “A” register (Accumulator).
  • The next three bits “000” – Immediate addressing mode.
• The binary equivalent of the operand “00 48” is “0000 0000 0100 1000”, which will be stored in the operand specifier.
• Since the addressing mode is immediate, the first byte of the operand specifier is ignored and the second byte value will be stored into the accumulator.
  • So, the second byte value “0100 1000” will be stored into the accumulator.

Thus, after the execution of the first instruction, the “A” register will contain the value “0100 1000”...