Answer
In this case would be address 11.
Work Step by Step
The value retrieved by a pop instruction would be the value at the current stack pointer, which is \( \mathrm{~A} \) (at address 12). After the pop operation, the stack pointer would point to the next value in the stack, which in this case would be address 11.
Given table:
Address Contents
10 F
11 C
12 A
13 B
14 E
Given:
Base of the stack = address 10
Stack pointer (SP) = 12
So:
Value popped:
→ Value at address 12 = A
Update stack pointer:
→ SP = SP − 1 = 12 − 1 = 11
Final Answers
Question Answer
Value retrieved by POP A
New stack pointer value 11
Summary of the stack after the pop:
Address Contents Comment
10 F still in stack
11 C now top of stack
12 A removed
13 B unused (above stack)
14 E unused (above stack)
So the new top of stack is now C (at address 11).
