Question 1. Consider the following Assembly program segment given below:

ORG 0x0400
Address
0400h MOV SP, 0460H
MOV AL, 1
MOV BX, 0
MOV CX, 7
AGAIN: CALL SHERIDAN
0420H SHL AL, 1
OUT 0, AL
LOOP AGAIN
INT3
SHERIDAN:
PUSH CX
MOV CX, AABBH
LOOP $
POP CX
RET
In executing the above program, the Microprocessor is using the stack. In that respect, draw the stack along with its content for the duration of the program execution.

3 answers

Based on the given program segment, the stack will be used to store the return address and the value of CX register. Here is the stack along with its content for the duration of the program execution:

Initial Stack:
| Address | Value |
|---------|-------|
| 0460 | 0460 |
| 045E | AABB |
|---------|-------|

Step 1:
| Address | Value |
|---------|-------|
| 045C | AA |
| 045A | BB |
| 0460 | 0460 |
| 045E | AABB |
|---------|-------|

Step 2:
| Address | Value |
|---------|-------|
| 045E | AA |
| 045C | BB |
| 0460 | 0460 |
|---------|-------|

Step 3:
| Address | Value |
|---------|-------|
| 045E | AA |
| 045C | BB |
| 0460 | 0460 |
| 045C | 07 |
|---------|-------|

Step 4:
| Address | Value |
|---------|-------|
| 045E | AA |
| 045C | BB |
| 0460 | 0460 |
| 045C | 07 |
| 045A | 42 |
|---------|-------|

Step 5:
| Address | Value |
|---------|-------|
| 045E | AA |
| 045C | BB |
| 0460 | 0460 |
| 045C | 07 |
| 045A | 42 |
| 0458 | 01 |
|---------|-------|

Step 6:
| Address | Value |
|---------|-------|
| 045E | AA |
| 045C | BB |
| 0460 | 0460 |
| 045C | 07 |
| 045A | 42 |
| 0458 | 01 |
| 0456 | 00 |
|---------|-------|
Can you explain each step?
Certainly! Here is an explanation of each step:

Step 1:

- MOV SP, 0460H: This instruction sets the stack pointer (SP) to the address 0460H.
- MOV AL, 1: This instruction moves the value 1 into the AL register.
- MOV BX, 0: This instruction moves the value 0 into the BX register.
- MOV CX, 7: This instruction moves the value 7 into the CX register.

Step 2:

- CALL SHERIDAN: This instruction calls the subroutine named SHERIDAN.
- PUSH CX: This instruction pushes the value of the CX register onto the stack.
- MOV CX, AABBH: This instruction moves the value AABBH into the CX register.
- LOOP $: This instruction decrements the CX register by 1 and jumps back to the address of the "LOOP $" instruction if CX is not zero. In this case, CX is not zero, so it jumps back to itself.
- POP CX: This instruction pops the value from the top of the stack into the CX register.
- RET: This instruction returns control back to the instruction following the CALL instruction.

Step 3:

- SHL AL, 1: This instruction shifts the bits in the AL register one position to the left.
- OUT 0, AL: This instruction outputs the value in the AL register to the address 0.
- LOOP AGAIN: This instruction decrements the CX register by 1 and jumps back to the address of the "LOOP AGAIN" instruction if CX is not zero. In this case, CX is 6, so it jumps back to the label AGAIN.

Step 4:

- LOOP $: This instruction decrements the CX register by 1 and jumps back to the address of the "LOOP $" instruction if CX is not zero. In this case, CX is 5, so it jumps back to itself.

Step 5:

- LOOP $: This instruction decrements the CX register by 1 and jumps back to the address of the "LOOP $" instruction if CX is not zero. In this case, CX is 4, so it jumps back to itself.

Step 6:

- LOOP $: This instruction decrements the CX register by 1 and jumps back to the address of the "LOOP $" instruction if CX is not zero. In this case, CX is 3, so it jumps back to itself.

Note: The remaining steps continue the same loop until the value of CX reaches zero.