In the Intel 8085 microprocessor, how is the hardware stack implemented in memory?

Introduction / Context:
The concept of a stack is central to the operation of microprocessors, especially for subroutine calls, interrupt handling and temporary storage of registers. The Intel 8085 uses a memory based stack that is controlled by a dedicated stack pointer register. Understanding how this stack is organised in memory helps when reading assembly code, debugging programs and designing interrupt service routines.

Given Data / Assumptions:

We are considering the Intel 8085 microprocessor architecture.
The 8085 has a 16 bit stack pointer register named SP.
The stack is used for PUSH, POP, CALL, RET and interrupt servicing.
The question asks how the hardware stack is implemented and how it grows in memory.
We assume a basic understanding of LIFO (last in first out) behaviour.

Concept / Approach:
In the 8085, the stack is not a separate hardware structure inside the CPU. Instead, it is a region of read write memory selected by the programmer. The 16 bit stack pointer register holds the address of the top of the stack. The stack operates in a last in first out manner. Importantly, the 8085 stack grows toward lower memory addresses. When data is pushed onto the stack, the stack pointer is decremented before writing the data. When data is popped, the data is read from the address pointed to by the stack pointer and then the stack pointer is incremented. This down growing memory based stack is typical of many eight bit processors.

Step-by-Step Solution:
Step 1: Recall that a LIFO stack removes items in the reverse order from which they were added. The top of the stack is where the most recent item resides. Step 2: In the 8085, the programmer initialises SP to some high memory address so that the stack has room to grow downward into unused memory. Step 3: When executing a PUSH instruction, the 8085 decrements SP by 2 because it pushes 16 bit register pairs, and then stores the data at the new address pointed to by SP. Step 4: When executing a POP instruction, the 8085 reads the 16 bit data from the address pointed to by SP and then increments SP by 2 to move the top of stack pointer back up. Step 5: This behaviour clearly shows that the stack is memory based, uses a stack pointer and grows toward lower addresses, making option A correct.

Verification / Alternative check:
If you step through a CALL instruction on an 8085 using a debugger, you will see that the return address is automatically pushed onto the stack. This involves decrementing the stack pointer and storing the program counter contents in memory at that location. Later, when RET is executed, the address is popped from the stack and SP is incremented. Examining the memory contents and SP values during this process confirms that the stack lives in memory and grows downward as items are pushed.

Why Other Options Are Wrong:
A built in hardware stack of fixed depth inside the CPU, with no use of general memory, is a feature of some microcontrollers, but it is not how the 8085 stack works.
A queue structure that grows toward higher addresses would implement FIFO behaviour, not the LIFO behaviour required for subroutines and interrupts.
A circular buffer stored only in internal registers without a stack pointer does not match the 8085 documentation and would be too limited for flexible program use.

Common Pitfalls:
A common mistake is to forget that the 8085 stack grows downward, which can lead to overlapping the stack with other data in memory if the starting address is chosen incorrectly. Another pitfall is failing to balance PUSH and POP operations, which corrupts the return address and causes program crashes. Programmers should always plan the stack region and ensure that stack operations remain consistent.

Final Answer:
The 8085 uses a memory based LIFO stack that grows toward lower addresses, managed by a 16 bit stack pointer register.