Stop execution Load the operand into the A register Store the contents of the A

Question

Stop execution Load the operand into the A register Store the contents of the A register into the operand Memory Value (in 1100 0001 0002 Add the operand to the A register CE Subtract the operand from the A register Character input to the operand Character output from the operand 1000 0004 0005 0006 2D 01010 01 FIGURE 64 Subset of Pep/8 instructions (a) What are the contents of the A register after the execution of this instruction? C1 00 01 b) what are the contents of the A register after the execution of this instruction? C1 00 03 (c ) What are the contents of the A register after the execution of the following two instructions? Co 00 05 70 00 05 (d) What are the contents of the A register after the execution of the following two instructions? c1 00 05 Eo 00 01 [5 Points] What are the 2 main parts of a Pep/8 machine language instruction? and (2) 5 Points] How many bits are there in a Pep/8 machine language instruction? xperience was produced by Dr. Alfred R. Watkins, Morehouse College -Spring

Explanation / Answer

Solution (a)

The given instruction is C1 00 01

Now C1 is 1100 0001. the Bit shown in bold signifies that register A will be used. and the LSB is 1 we load the contents at the address 0001. Which is 1A

Thus contents of register A is 1A

Soulution(b):

Same explaination as above. Instruction given C1 00 03. So contents of A are 55

Solution(c):

The first instruciotn is C0 00 05. As the LSB in opcode C0 is 0 we will copy 0005 into register A

The second instruction is 70 00 05. As here again the LSB is 0 we consider the given number as immediate. Hence we have A = A + 0005. As A = 0005 from first instruction. Contents of A will be 000A

Solution (d):

The first instruction is C1 00 05. LSB is 1 so contents at location 0005 which are 00. Hence A = 00

Second instruction E0 00 01.

According to the given link

http://suffolk.li/cst111/19cst111/pep8.html

The opcode E0 is illegal instructions. Either your prof has given it purposefully or you better check with your prof about this.

Solution (e)

2 Main parts of PEP/8 instructions:

1. one byte instruction specifier

2. 16 bit operand.

solution(f)

Number of bits in PEP/8 language.

there are total 24 bits in an instruction.

Of which first 4 bits are opcode. In 2 nibble the first bit specifies if register A or X and the last three bits specifies immediate or memory. The remaining 16 bits are operand.

Thanks.

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