Hello, this is a 2 part question for my beginning assembly class. Thank you in advance for your help

Boolean Calculator (1)

Create a program that functions as a simple boolean calculator for 32-bit integers. It should display a menu that asks the user to make a selection from the following list:

1. x AND y

2. x OR y

3. NOT x

4. x XOR y

5. Exit Program

When the user makes a choice, call a procedure that displays the name of the operation about to be performed.

Boolean Calculator (2)

Continue the solution program from the preceeding exercise by implementing the following procedures:

- AND_op: Prompt the user for two hexadecimal integers. AND them together and display the result in hexadecimal.

-OR_op: Prompt the user for two hexadecimal integers. OR them together and display the result in hexadecimal.

- NOT_op: Prompt the user for two hexadecimal integers. NOT the integer and display the result in hexadecimal.

- XOR_op: Prompt the user for two hexadecimal integers. Exclusive-OR them together and display the result in hexadecimal.

The following is what our professor has given us help with:

Include Irvine32.inc

.data
msgMenu " Boolean Calculator ", 0dh, 0ah
BYTE 0dh, 0ah
BYTE "1. x AND y" , 0dh, 0ah
BYTE "2. x OR y" , 0dh, 0ah
BYTE "3. NOT x" , 0dh, 0ah
BYTE "4. x XOR y" , 0dh, 0ah
BYTE "5. Exit program" , 0dh, 0ah, 0dh, 0ah
BYTE "Enter integer>" , 0

msgAND BYTE "Boolean AND", 0
msgOR BYTE "Boolean OR", 0
msgNOT BYTE "Boolean NOT", 0
msgXOR BYTE "Boolean XOR", 0

msgOperand1 BYTE "Input the first 32--bit hexadecimal operand: ", 0
msgOperand2 BYTE "Input the second 32--bit hexadecimal operand: ", 0
msgResult BYTE "The 32-bit hexadecimal result is: ", 0

Complete code is given below:

Explanation:

INCLUDE Irvine32.inc

.data

msgMenu BYTE " Boolean Calculator ", 0dh,0ah

BYTE 0dh,0ah

BYTE "1. x AND y"     ,0dh,0ah

BYTE "2. x OR y"      ,0dh,0ah

BYTE "3. NOT x"       ,0dh,0ah

BYTE "4. x XOR y"     ,0dh,0ah

BYTE "5. Exit program",0

msgAND      BYTE "Boolean AND",0

msgOR       BYTE "Boolean OR",0

msgNOT      BYTE "Boolean NOT",0

msgXOR      BYTE "Boolean XOR",0

msgOperand1 BYTE "Input the first 32-bit hexadecimal operand: ",0

msgOperand2 BYTE "Input the second 32-bit hexadecimal operand: ",0

msgResult   BYTE "The 32-bit hexadecimal result is:            ",0

caseTable   BYTE '1'     # lookup value

DWORD _AND_op        # address of the procedure

EntrySize = ($ - caseTable )

   BYTE '2'

   DWORD _OR_op

   BYTE '3'

   DWORD _NOT_op

   BYTE '4'

   DWORD _XOR_op

   BYTE '5'

   DWORD _ExitProgram

NumberOfEntries=($ - caseTable)/EntrySize

.code

main08stub PROC

   call Clrscr

Menu:

   mov edx, OFFSET msgMenu # menu choices

   call WriteString

   call Crlf

L1: call ReadChar

   cmp al, '5' # is selection valid (1-5)?

   ja L1   # if above 5, go back

   cmp al, '1'

   jb L1   # if below 1, go back

   call Crlf

   call **BLANK**

   jc quit # if CF = 1 exit

   call Crlf

   jmp Menu    # display menu again

quit: exit

main08stub ENDP

#

**BLANK** PROC

#

# It selects a procedure from the case table

# Receives: AL = number of procedure

# Returns: nothing

#

   ret

**BLANK** ENDP

#

_AND_op PROC

#

# Performs a boolean AND operation

# Receives: Nothing

# Returns: Nothing

#

   ret

_AND_op ENDP

#

_OR_op PROC

#

# Performs a boolean OR operation

# Receives: Nothing

# Returns: Nothing

#

   ret

_OR_op ENDP

#

_NOT_op PROC

#

# Performs a boolean NOT operation.

# Receives: Nothing

# Returns: Nothing

#

   ret

_NOT_op ENDP

#

_XOR_op PROC

#

#

# Performs an Exclusive-OR operation

# Receives: Nothing

# Returns: Nothing

#

   ret

_XOR_op ENDP

#

_ExitProgram PROC

#

# Receives: Nothing

# Returns: Sets CF = 1 to signal end of program

#

   ret

_ExitProgram ENDP

END main08stub

I get 0x55 and this the linking address of the main function.

use this function to see changes:

/* bar6.c */

#include <stdio.h>

char main1;

void p2()

{

printf("0x%X\n", main1);

}

Output is probably 0x0

you can use your original bar6.c with updaated foo.c

char main;

int main() // error because main is already declared

{

  p2();

   //printf("Main address is 0x%x\n",main);

  return 0;

}

Will give u an error

again

int main()

{

  char ch = main;

  p2(); //some value

  printf("Main address is 0x%x\n",main); //some 8 digit number not what printed in p2()

  printf("Char value is 0x%x\n",ch); //last two digit of previous line output

  return 0;

}

So the pain in P2() gets the linking address of the main function and it is different from address of the function main.

Now char main (uninitialized) in another compilation unit fools the compiler by memory-mapping a function pointer on a char directly, without any conversion: that's undefined behavior. Try char main=12; you'll get a multiply defined symbol main...

Explanation: