1. The Question: The question asks for the final value of the stack pointer (SP) after the CALL instruction executes.
2. What the CALL Instruction Does (to the Stack)
The problem explicitly tells us the CALL implementation:
Store the current value of PC in the stack.
Store the value of PSW register in the stack.
(Load the new address into PC - this does not affect the stack).
So, the only thing that happens to the stack is that two items are pushed onto it: first the PC, then the PSW.
3. Key Information to Solve the Problem
Current SP: The stack pointer starts at $(016E)_{16}$.
Item Size:
Stack Behavior (The Most Important Part):
4. Step-by-Step Calculation
This is where we trace the SP.
What "Grows Upward" Means: Most stacks in computers grow downward (pushing decreases the address). This stack is different. "Grows upward" means that every time you push an item, the stack pointer's address increases.
What "SP points to the top element" Means: The SP's current value, $(016E)_{16}$, is the address of the last item currently on the stack.
Let's trace the operation:
Initial State:
Operation 1: Store PC (2 bytes)
We need to push 2 bytes onto the stack.
Because the stack grows upward, we add the size to the SP.
New SP = (Old SP) + (Size of PC)
New SP = $(016E)_{16} + 2$
0 1 6 E (in hex)
+ 2 (in hex)
----------
In hexadecimal, the letter E represents the decimal number 14.
So, this is $14 + 2 = 16$.
How do we write 16 in hexadecimal? It's 10. (It's one "16" and zero "1s").
This means we write down 0 and carry over the 1 to the next column.
1
0 1 6 E
+ 2
----------
0
1
0 1 6 E
+ 2
----------
7 0
0 1 6 E
+ 2
----------
0 1 7 0
Operation 2: Store PSW (2 bytes)
Next, we push the 2-byte PSW onto the stack.
New SP = (Current SP) + (Size of PSW)
New SP = $(0170)_{16} + 2$
New SP = $(0172)_{16}$
(The 2 bytes of the PSW are stored at $(0171)_{16}$ and $(0172)_{16}$. The SP now points to the final top element at $(0172)_{16}$).
Final Value: $(0172)_{16}$
"Trap" Information (What to Ignore)
This problem is difficult because it gives you three pieces of information that are completely irrelevant to finding the final stack pointer value.
TRAP 1: The content of PC just before the fetch... is $(5FA0)_{16}$
Why it's a trap: This information is needed to calculate the return address (the value of the PC that gets pushed). The CALL instruction is 2 words (4 bytes) long, so the return address pushed would be $(5FA0)_{16} + 4 = (5FA4)_{16}$. But we don't care what value is pushed; we only care that 2 bytes are pushed.
TRAP 2: The CALL instruction is of two words...
TRAP 3: A stack... is implemented from memory location $(0100)_{16}$
The most common mistake is to assume the stack grows downward, which would lead to $(016E)_{16} - 4 = (016A)_{16}$ (Option A).
