Chapter 6, Problem 34E

Explanation of Solution

Execution of the first instruction “0001 31 00 0F”:

Given instruction, 0001 31 00 0F

• The value “0001” is the address of the instruction.
• The hexadecimal value “31” is an instruction and “00 0F” is the operand.
• The mnemonic for the hexadecimal value “31” is “DECI”.
• The binary value for the hexadecimal instruction “31” is “0011 0001”.
• The first five bits express the meaning of the instruction.
  • “00110” – Decimal input to the operand “(DECI)”.
• So, the instruction “31” gets the decimal input from the user.
  • Enter “5” as input.
• The last three bits are used as addressing mode specifier.
  • “001” – Direct addressing mode.
    • The direct addressing mode represents the address of data in the operand.
• So, it loads the input value “5” to the given operand’s location “00 0F”.

Address and its values

Address	Value
000F	5

Execution of the second instruction “0004 C1 00 0F”:

Given instruction, 0004 C1 00 0F

• The value “0004” is the address of the instruction.
• The hexadecimal value “C1” is an instruction, and “00 0F” is the operand.
• The mnemonic for the hexadecimal value “C1” is “LDr”.
• The binary value for the hexadecimal instruction “C1” is “1100 0001”.
• The first four bits express the meaning of the instruction.
  • “1100” – Load the operand into the A register “(LDr)”.
• The last three bits are used as addressing mode specifier.
  • “001” – Direct addressing mode.
    • The direct addressing mode represents the address of data in the operand.
    • Here, the value in the address represented in the operand “00 0F” is “5”.
• So, it loads the value “5”, from the address stored in the operand “00 0F”, into the accumulator...