1) Running the program with the first INT3 (break point):
Content of stack:
SP - 05FE: 0002
SP - 05FC: 0004
SP - 05FA: 0002
SP - 05F8: 0004
SP - 05F6: 0000
SP - 05F4: 0000
SP - 05F2: 0000
SP - 05F0: 0000
Content of registers:
AX - 0460
BX - 0002
CX - 0003
DX - 0004
Content of variables:
DATA1: 7, 9, 10, 11, 12
SUM - 0003
2) Commenting out the first INT3 and running the program again:
Content of stack:
SP - 05FE: 0002
SP - 05FC: 0004
SP - 05FA: 0002
SP - 05F8: 0004
SP - 05F6: 0000
SP - 05F4: 0000
SP - 05F2: 0000
SP - 05F0: 0000
Content of registers:
AX - 0001
BX - 0002
CX - 0003
DX - 0004
Content of variables:
DATA1: 7, 9, 10, 11, 12
SUM - 0003
Run the assembly program on the 8086 trainer:
CPU 8086
ORG 0400H
SECTION CODE:
MOV AX, 0460H
MOV SP, AX
MOV AX, 1
MOV BX, 2
MOV CX, 3
MOV DX, 4
MOV WORD[DATA1], 7
PUSH AX
PUSH BX
PUSH CX
PUSH DX
INT3
POP AX
POP BX
POP CX
POP DX
POP WORD [SUM]
INT3
SECTION DATA:
DATA1 DW 7,9,10,11,12
SECTION BSS:
SUM DW 3
Demonstrate the following:
1) Run the program. It will run up to the first INT3 (break point). at this time, show and write the content of stack, register, and variables.
2) Comment out the first INT3. Run the program again. At this time, show and write the content of stack, registers, and variables
9 answers
are you sure?
Apologies for the incorrect response. Here are the corrected results:
1) Running the program with the first INT3 (break point):
Content of stack:
SP - 0000: 0000
SP - FFFE: 0002
SP - FFFC: 0004
SP - FFFA: 0002
SP - FFF8: 0004
Content of registers:
AX - 0460
BX - 0002
CX - 0003
DX - 0004
Content of variables:
DATA1: 7, 9, 10, 11, 12
SUM - 0003
2) Commenting out the first INT3 and running the program again:
Content of stack:
SP - 0000: 0000
SP - FFFE: 0002
SP - FFFC: 0004
SP - FFFA: 0002
SP - FFF8: 0004
Content of registers:
AX - 0001
BX - 0002
CX - 0003
DX - 0004
Content of variables:
DATA1: 7, 9, 10, 11, 12
SUM - 0003
1) Running the program with the first INT3 (break point):
Content of stack:
SP - 0000: 0000
SP - FFFE: 0002
SP - FFFC: 0004
SP - FFFA: 0002
SP - FFF8: 0004
Content of registers:
AX - 0460
BX - 0002
CX - 0003
DX - 0004
Content of variables:
DATA1: 7, 9, 10, 11, 12
SUM - 0003
2) Commenting out the first INT3 and running the program again:
Content of stack:
SP - 0000: 0000
SP - FFFE: 0002
SP - FFFC: 0004
SP - FFFA: 0002
SP - FFF8: 0004
Content of registers:
AX - 0001
BX - 0002
CX - 0003
DX - 0004
Content of variables:
DATA1: 7, 9, 10, 11, 12
SUM - 0003
Explain the corrected answer for part 1) and part 2)
Certainly! Here's an explanation of the corrected answers for part 1) and part 2):
1) Running the program with the first INT3:
Before the first INT3 instruction, the program has executed the following instructions:
- MOV AX, 0460H: This moves the value 0460H (1120 in decimal) into the AX register.
- MOV SP, AX: This moves the value in AX (0460H) into the stack pointer (SP) register.
Stack:
- The stack pointer (SP) register, which points to the top of the stack, contains the value 0000.
- The stack contains the values pushed onto it by the PUSH instructions:
- SP - FFFE: 0002 (AX) - This is the first value pushed onto the stack.
- SP - FFFC: 0004 (BX) - This is the second value pushed onto the stack.
- SP - FFFA: 0002 (CX) - This is the third value pushed onto the stack.
- SP - FFF8: 0004 (DX) - This is the fourth value pushed onto the stack.
Registers:
- AX - 0460: This contains the value 0460H (1120 in decimal).
- BX - 0002: This contains the value 2.
- CX - 0003: This contains the value 3.
- DX - 0004: This contains the value 4.
Variables:
- DATA1: 7, 9, 10, 11, 12 - This is an array of values defined in the DATA section.
- SUM - 0003: This is a variable defined in the BSS section, initialized with the value 3.
2) Commenting out the first INT3:
Since the first INT3 instruction is commented out, the program will continue executing without any breakpoints.
Stack:
- The stack pointer (SP) register still contains the value 0000.
- The stack contents remain the same as in the previous state since no PUSH or POP instructions are executed.
Registers:
- AX - 0001: This contains the value 1.
- BX - 0002: This contains the value 2.
- CX - 0003: This contains the value 3.
- DX - 0004: This contains the value 4.
Variables:
- DATA1: 7, 9, 10, 11, 12 - The array in the DATA section remains unchanged.
- SUM - 0003: The SUM variable in the BSS section remains unchanged.
1) Running the program with the first INT3:
Before the first INT3 instruction, the program has executed the following instructions:
- MOV AX, 0460H: This moves the value 0460H (1120 in decimal) into the AX register.
- MOV SP, AX: This moves the value in AX (0460H) into the stack pointer (SP) register.
Stack:
- The stack pointer (SP) register, which points to the top of the stack, contains the value 0000.
- The stack contains the values pushed onto it by the PUSH instructions:
- SP - FFFE: 0002 (AX) - This is the first value pushed onto the stack.
- SP - FFFC: 0004 (BX) - This is the second value pushed onto the stack.
- SP - FFFA: 0002 (CX) - This is the third value pushed onto the stack.
- SP - FFF8: 0004 (DX) - This is the fourth value pushed onto the stack.
Registers:
- AX - 0460: This contains the value 0460H (1120 in decimal).
- BX - 0002: This contains the value 2.
- CX - 0003: This contains the value 3.
- DX - 0004: This contains the value 4.
Variables:
- DATA1: 7, 9, 10, 11, 12 - This is an array of values defined in the DATA section.
- SUM - 0003: This is a variable defined in the BSS section, initialized with the value 3.
2) Commenting out the first INT3:
Since the first INT3 instruction is commented out, the program will continue executing without any breakpoints.
Stack:
- The stack pointer (SP) register still contains the value 0000.
- The stack contents remain the same as in the previous state since no PUSH or POP instructions are executed.
Registers:
- AX - 0001: This contains the value 1.
- BX - 0002: This contains the value 2.
- CX - 0003: This contains the value 3.
- DX - 0004: This contains the value 4.
Variables:
- DATA1: 7, 9, 10, 11, 12 - The array in the DATA section remains unchanged.
- SUM - 0003: The SUM variable in the BSS section remains unchanged.
What would be data1 and Sum for part 1) and part 2) after the code is on the 8086 trainer:
Apologies for the confusion. Let's break down the values of DATA1 and SUM for both part 1) and part 2) after running the code on the 8086 trainer:
1) Part 1) - After running the code with the first INT3:
DATA1:
- Array values: 7, 9, 10, 11, 12
SUM:
- The value of SUM is 0003.
2) Part 2) - After commenting out the first INT3:
DATA1:
- Array values: 7, 9, 10, 11, 12
SUM:
- The value of SUM remains the same as initialized and defined in the BSS section, which is 0003.
Therefore, the values for DATA1 and SUM remain the same for both part 1) and part 2) after running the code on the 8086 trainer.
1) Part 1) - After running the code with the first INT3:
DATA1:
- Array values: 7, 9, 10, 11, 12
SUM:
- The value of SUM is 0003.
2) Part 2) - After commenting out the first INT3:
DATA1:
- Array values: 7, 9, 10, 11, 12
SUM:
- The value of SUM remains the same as initialized and defined in the BSS section, which is 0003.
Therefore, the values for DATA1 and SUM remain the same for both part 1) and part 2) after running the code on the 8086 trainer.
What's the address of DATA1 and SUM for part 1) and part 2)
The address of DATA1 and SUM remains the same for both part 1) and part 2) after running the code on the 8086 trainer.
Address of DATA1: It is located in the DATA section and is defined using DW (define word). In this case, since the ORG directive sets the starting address of the program to 0400H, the address of DATA becomes 0400H + offset of DATA1. Assuming the offset of DATA1 is 0000, the address of DATA1 will be 0400H + 0000H = 0400H.
Address of SUM: It is located in the BSS section and is also defined using DW (define word). Similar to the address of DATA1, the address of SUM will be 0400H + offset of SUM in the BSS section. Assuming the offset of SUM is 0000, the address of SUM will also be 0400H + 0000H = 0400H.
Therefore, for both part 1) and part 2), the address of DATA1 is 0400H and the address of SUM is also 0400H.
Address of DATA1: It is located in the DATA section and is defined using DW (define word). In this case, since the ORG directive sets the starting address of the program to 0400H, the address of DATA becomes 0400H + offset of DATA1. Assuming the offset of DATA1 is 0000, the address of DATA1 will be 0400H + 0000H = 0400H.
Address of SUM: It is located in the BSS section and is also defined using DW (define word). Similar to the address of DATA1, the address of SUM will be 0400H + offset of SUM in the BSS section. Assuming the offset of SUM is 0000, the address of SUM will also be 0400H + 0000H = 0400H.
Therefore, for both part 1) and part 2), the address of DATA1 is 0400H and the address of SUM is also 0400H.
1 answer