ANSWER QUESTIONS 13-24 Question 14 Meaning of Instruction : A)Stop Execution B)L

Question

ANSWER QUESTIONS 13-24

Question 14

Meaning of Instruction :

A)Stop Execution

B)Load the operand into the A register

C)Store the contents of the A register into the operand

D)Add the operand to the A register

E)Subtract the operand from the A register

F)Character input to the operand

G)Character output to the operand

Question 15

Addressing Mode:

A)Immediate

B)Direct

Question 16

Operand Specifier (2-byte word):

A) 11 00

B) AB CD

C) C0

D) 11

E) 00 02

Question 17

Operand (2-byte word):

A) 11 00

B) 00 02

C) 11

D) AB CD

E) E1

Question 18

Contents of the A register after execution (2-byte word) :

A) AB CF

B) 11 02

C) 00 02

D) AB CD

E) 11 00

Question 19

Instruction Specifier (Binary) :__________________________

Question 20 (1 point)

Meaning of Instruction :

A)Stop Execution

B)Load the operand into the A register

C)Store the contents of the A register into the operand

D) Add the operand to the A register

E) Subtract the operand from the A register

F) Character input to the operand

G) Character output to the operand

Question 21

Addressing Mode:

A) Immediate

B) Direct

Question 22

Operand Specifier (2-byte word):

A) 00 00

B) 00

C) AB CD

D) 81

E) 00 01

Question 23

Operand (2-byte word):

A) 81

B) 00 00

C) 00 01

D) 00

E) AB CD

Question 24

Contents of the A register after execution (2-byte word) :

A) AB CC

B) 00 00

C) 00 01

D) AB CD

A)Stop Execution

B)Load the operand into the A register

C)Store the contents of the A register into the operand

D)Add the operand to the A register

E)Subtract the operand from the A register

F)Character input to the operand

G)Character output to the operand

Information Given the following Pep/8 virtual computer state of memory (in hexadecimal RAM Address Content 0000 00 0001 01 0002 0003 00 0004 and the machine instruction 70 00 02. Note: When the instruction is executed, Reg A contains the value AB CD. Question 13 (5 points) d Instruction Specifier (Binary):

Explanation / Answer

Answers:

Instruction is 70 00 02.

Reg A contains AB CD.

(13). Instruction specifier is 70 which is in hexadecimal
So in binary it is 01110000

(14). This instruction is ADD A,i which means add the immediate operand to register A.

So answer is option (D).

(15). This is immediate addressing mode (from opcode 70) so answer is option (A).

(16). Operand specifier is 00 02 so the answer is (E).

(17). As this is the immediate addressing mode, the operand specifier itself is an operand. So operand is 00 02. Which means the answer is (B).

(18). After execution , A register contains AB CF. Before execution, it was having AB CD and 00 02 is added into it. ABCD + 0002 = ABCF. So, the answer is option (A).

------------------------------------------------

Instruction is 81 00 00. Register A contains AB CD.

(19). Instruction specifier is 81 in hexadecimal which is in binary 10000001.

(20). This instruction is SUB A, d which means subtract Accumulator with the memory content and store the result in register A.
So the correct answer is (E).

(21). Addressing mode is direct as it is also getting meaning from 81 SUB A, d and here d is for direct. So the answer is (B).

(22). Operand specifier is 00 00 so the correct option is (A).

(23). As this instruction is dealing with memory, the operand will be stored at the location specified by operand specifier which means at memory location 0000. As per the given data, the content at address 0000 is 00. And as this is instruction of 2-byte word, the operand is also of 2-byte. Location 0000 gives the first byte and so the next byte data is taken from address 0001 which is 01. This means the content of 0000 and 0001 memory locations which is 00 01 is the operand. So the answer is (C).

(24). A register content was AB CD. Now, after subtracting 00 01 from it, the register content becomes AB CC as AB CD - 00 01 = AB CC. So the answer is (A).

Please comment if there is any query. Thank you. :)

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