Question

Convert the following pep/9 machine language program into an assembly code. Determine the output of this program if the input is ‘tab’. The left column is the memory address of the first byte on the line:

0000 D1FC15

0003 F1001F

0006 D1FC15

0009 F10020

000C D1FC15

000F F10021

0012 D10020

0015 F1FC16

0018 D1001F

001B F1FC16

001E 00

Answer 1

Address(in hex.)	Instruction in hex.	Instruction in binary	Meaning of the instructions
0000	D1FC15	1101 0001 1111 1100 0001 0101	Instruction = 1101 0001 1101 = Load byte r(8..15) from memory (1 byte from memory referenced by the operand to right half of the accumulator) raaa=0001. r=0(means load to accumulator) aaa=001(means direct addressing) operand=1111 1100 0001 0101 FC15=refers input device(keyboard) Load a byte(character ‘t’) from key board
0003	F1001F	1111 0001 0000 0000 0001 1111	Instruction = 1111 0001 1111 = Store byte r(8..15) to memory (Store the right half of the accumulator(1 byte) to memory referenced by the operand) raaa=0001. r=0(means store from accumulator) aaa=001(means direct addressing) operand= 0000 0000 0001 1111 Store(print) byte stored in the right half of the accumulator into memory address 001F. Since recently ‘t’ is stored in the right half of the accumulator, ‘t’ will be stored at memory address 001F.
0006	D1FC15	1101 0001 1111 1100 0001 0101	Load a byte(character ‘a’) from key board
0009	F10020	1111 0001 0000 0000 0010 0000	Store(print) byte stored in the right half of the accumulator into memory address 0020. Since recently ‘a’ is stored in the right half of the accumulator, ‘a’ will be stored at memory address 0020.
000C	D1FC15	1101 0001 1111 1100 0001 0101	Load a byte(character ‘b’) from key board
000F	F10021	1111 0001 0000 0000 0010 0001	Store(print) byte stored in the right half of the accumulator into memory address 0021. Since recently ‘b’ is stored in the right half of the accumulator, ‘b’ will be stored at memory address 0021.
0012	D10020	1101 0001 0000 0000 0010 0000	Load a byte from memory address 0020 Since Byte(character) stored at 0020 is ‘a’, ‘a’ will be stored into the right half of the accumulator.
0015	F1FC16	1111 0001 1111 1100 0001 0110	FC16=refers output device(console or screen) Store(print) byte stored in the right half of the accumulator to screen. Since recently ‘a’ is stored in the right half of the accumulator, ‘a’ will be printed to the screen.
0018	D1001F	1101 0001 0000 0000 0001 1111	Load a byte from memory address 001F Since Byte(character) stored at 001F is ‘t’, ‘t’ will be stored into the right half of the accumulator.
001B	F1FC16	1111 0001 1111 1100 0001 0110	FC16=refers output device(console or screen) Store(print) byte stored in the right half of the accumulator to the screen. Since recently ‘t’ is stored in the right half of the accumulator, ‘t’ will be printed to the screen.
001E	00	0000 0000	Stop execution

Input: tab

Output: at

Therefore, the given pep/9 program takes input 'tab' and prints 'at'.