Your question:

PROGRAM 7: In-n-Out!

for hla 176 bat compiler

Below are some sample program dialogues that demonstrate these ideas.
(Hint: Do this in small steps, bit-by-bit. There's alot to it... )
(Another Hint: HLA read in hex format when you read directly into a register. So do that...)
(Further Hint: The most important part of this assignment is to worked with the packed data field entered by the user to extract the sub-parts out of it. The overlapping design of the Intel registers helps you parse this kind of data field and you can shift the bits around to get the right part into BX or CX, for example... )
(Final Hint: Since we haven't learned how to do multiplication yet, although it's kinda painful, I was expecting that you would perform the multiplication by a looping set of addition instructions)

Feed me your order as 4 hex digits: 0001
0 Combo 1s
0 Combo 2s
1 Combo 3s
Total Order Costs: $6

Feed me 4 hex digits: 0021
0 Combo 1s
1 Combo 2s
1 Combo 3s
Total Order Costs: $14

Feed me 4 hex digits: 04C9
1 Combo 1s
6 Combo 2s
9 Combo 3s
Total Order Costs: $112

Feed me 4 hex digits: 0009
0 Combo 1s
0 Combo 2s
9 Combo 3s
Total Order Costs: $54

Assignment Help Answers with Step-by-Step Explanation:

combo1Cost: dword := 10; // Cost of Combo 1

combo2Cost: dword := 8; // Cost of Combo 2

stdin.gethex(userInput);

// Extract the number of Combo 3 items (the last 5 bits)

shr(userInput, 5);

// Extract the number of Combo 2 items (the next 5 bits)

// The remaining bits are the number of Combo 1 items

mov(al, bl);

addLoop1:

cmp(ecx, bl); // Compare counter to the number of Combo 1 items

doneCombo1:

// Loop to calculate the cost for Combo 2

add(totalCost, combo2Cost); // Add the cost of Combo 2 to the total cost

inc(ecx); // Increment the counter

addLoop3:

cmp(ecx, cl); // Compare counter to the number of Combo 3 items

doneCombo3:

// Display the results

stdout.put(inttohex(dl, false));

stdout.newln();

stdout.put(totalCost, nl);

end InNOutOrder;