Vasily S. answered • 3d
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1. Setup and Initialization
This phase prepares the program to run.
Set Memory Model: **.MODEL SMALL** tells the assembler to use a simple memory layout: one segment (64KB) for code and one segment (64KB) for data.
Reserve Stack: **.STACK 100H** allocates 256 bytes for the stack. The stack is a temporary holding area, essential for CALL, PUSH, and POP instructions.
Define Data: The **.DATA** section reserves memory for all variables.
Prompts: prompt1, prompt2, prompt3, and result_msg store the literal text strings. Each string must end with a $ character. This $ is a terminator, telling the DOS print function (AH=09H) where to stop printing.
Input Buffer: input_buffer defines a structure needed for the DOS buffered input function (AH=0AH). It reserves space for the maximum length, the actual length, and the string data itself.
Binary Storage: num1, num2, and num3 are defined as 16-bit "words" (DW). These variables store the final binary numbers after they are converted from the user's text input.
2. Main Execution (MAIN PROC)
This is the central logic flow of the program.
Initialize Data Segment (DS):
MOV AX, @DATA
MOV DS, AX
This is the most critical setup step. The CPU's DS register must point to the .DATA section. Without these two lines, the program cannot find any variables and will crash.
Get First Number (A):
LEA DX, prompt1: Loads the memory address of the first prompt string into the DX register.
MOV AH, 09H / INT 21H: This is the DOS "Print String" interrupt. It prints the string pointed to by DX.
CALL READ_SIGNED_INT: This instruction pauses the MAIN procedure and jumps to the READ_SIGNED_INT procedure. The program waits there until the user types a number and presses ENTER.
When the procedure finishes, it returns control to this exact point. The converted binary number (e.g., 100) is now in the AX register.
MOV num1, AX: The program saves the value from AX into the num1 variable.
Get Second Number (B):
This process repeats exactly for the second number. It prints prompt2, calls the same READ_SIGNED_INT procedure, and stores the new result from AX into num2.
Get Third Number (C):
This process repeats again for the third number, storing the result from AX into num3.
Perform Calculation:
MOV AX, num1: The CPU loads the value of num1 (A) into the AX register, overwriting its previous content.
ADD AX, num2: The CPU adds the value of num2 (B) directly to the AX register. AX now holds the sum (A + B).
SUB AX, num3: The CPU subtracts the value of num3 (C) from the AX register.
AX now holds the final binary result of (A + B - C).
Display the Result:
LEA DX, result_msg: Loads the address of the "The result is: " string into DX.
MOV AH, 09H / INT 21H: Prints that string to the screen.
CALL PRINT_SIGNED_INT: The program jumps to the PRINT_SIGNED_INT procedure. This procedure takes the final answer (which is still sitting in AX from the calculation) and converts it back into a string of text to display.
Exit Program:
MOV AH, 4CH / INT 21H: This is the standard, clean DOS "Terminate Program" function. It stops the program and returns control to the command line.
3. Procedures (The Sub-Tasks)
These are the reusable "helper" functions.
READ_SIGNED_INT (Convert Text to Number)
This procedure converts a string of ASCII characters (like "1", "2", "5") into a single 16-bit binary number (125).
Read String: It uses AH=0AH / INT 21H to read the user's keystrokes into the input_buffer.
Setup Conversion: It zeroes out three registers: AX (to build the result), BX (to hold the current digit), and CX (to act as a sign flag).
Check for Sign: It inspects the first character of the input. If it is a '-' (minus sign), it sets CX to 1. This "flags" the number as negative.
Conversion Loop: The program iterates through the input string, one character at a time. For each character:
It multiplies the current total in AX by 10. (e.g., If AX holds 12, it becomes 120).
It converts the ASCII digit to a number by subtracting '0'. (e.g., The character '5' becomes the number 5).
It adds the new digit to the total. (e.g., 120 + 5 = 125).
This loop repeats until it hits a non-digit character (like the ENTER key).
Apply Sign: Before finishing, it checks the CX flag. If CX is 1 (meaning a '-' was found), it negates the entire number in AX using the NEG instruction.
Return: RET sends control back to MAIN. The final signed binary number is left in AX.
PRINT_SIGNED_INT (Convert Number to Text)
This procedure does the exact opposite. It converts a 16-bit binary number (like 125) into a string of ASCII characters ("1", "2", "5") and prints them.
Check for Sign: It checks if the number in AX is negative.
Handle Negative: If the number is negative, it first prints a '-' character to the screen. It then uses the NEG instruction to make the number in AX positive. This simplifies the following math.
Conversion Loop (Divide by 10): This algorithm breaks the number apart, digit by digit, from right to left.
DIV BX: Divides the number in AX by 10 (which is in BX).
The quotient (e.g., 12) goes back into AX.
The remainder (e.g., 5) goes into DX. This remainder is the last digit.
ADD DL, '0': Converts the remainder (5) into its ASCII character ('5').
PUSH DX: Pushes the character onto the stack.
This loop repeats. On the next pass, it divides 12 by 10 (getting 2), pushes '2'. Then it divides 1 by 10 (getting 1), pushes '1'. The loop stops when AX becomes 0.
The stack now holds the digits in reverse order: '1', '2', '5'.
Printing Loop:
POP DX: Pops a character from the stack. Because of the stack's "Last-In, First-Out" nature, the digits are popped in the correct order.
MOV AH, 02H / INT 21H: This is the DOS "Print Character" interrupt. It prints the single digit in DL.
LOOP: This command repeats the pop-and-print process until the stack is empty (CX counts the digits).
Return: RET sends control back to MAIN, and the program finishes.
