Big-endian storage and the stack: 22-Elec-A4 Digital Systems and Computers, December 2018, Q5

Question 5: Big-endian storage and the stack (12 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Big-endian (Motorola) byte order; a 16-bit value 0x7A01 (high byte 0x7A, low byte 0x01); target address 0xC239; stack pointer SP = 0xDC51 pointing at the next free byte.

Find. The bytes placed in memory for the store and the push, and the post-push value of SP.

Approach. Big-endian places the most-significant byte at the lowest address. A Motorola PUSH writes the low byte at the current SP, decrements, writes the high byte, and decrements again, so the two bytes still end up big-endian in memory with SP left below them.

(a) Store 0x7A01 → 0xC239. High byte first at the low address: address 0xC239 receives 0x7A and 0xC23A receives 0x01; 0xC238 is untouched.
Part (a): big-endian store — high byte 0x7A at 0xC239, low byte 0x01 at 0xC23A.
(b) Push 0x7A01 with SP = 0xDC51. The PUSH stacks the low byte first: $[\text{0xDC51}]\leftarrow$ 0x01, then SP decrements and $[\text{0xDC50}]\leftarrow$ 0x7A. In memory the high byte 0x7A sits at the lower address 0xDC50 — big-endian, consistent with part (a).
Part (b): after the push, 0xDC50 = 0x7A (high) and 0xDC51 = 0x01 (low).
(c) Stack pointer after PUSH. Two bytes were pushed, so SP decrements twice from 0xDC51: $$\text{SP}=\text{0xDC51}-2=\boxed{\mathtt{DC4F}_{16}}.$$ It again points at the next free (lower) byte, below the value just stored.

Question 5 — results
Item	Result
Store: 0xC239 / 0xC23A	0x7A / 0x01
Push: 0xDC50 / 0xDC51	0x7A / 0x01
SP after PUSH	0xDC4F

Topic. Endianness and stack operation on a Motorola-style CPU.

Key relations. Big-endian: MS byte → lowest address. Motorola PUSH of a 16-bit register: $[\,SP\,]\leftarrow$ low, $SP\!-\!1$, $[\,SP\,]\leftarrow$ high, $SP\!-\!1$ (stack grows toward low memory).

Why this works. Because the SP points at the next free byte and the low byte is stored first while the pointer moves downward, the high byte lands one address below the low byte — which is precisely big-endian ordering, matching how the same value is laid out by a plain store. The pointer ends two below its start because a 16-bit datum occupies two bytes.

Common pitfall. Writing the low byte at the low address (that is little-endian, the wrong convention here), or moving SP by only one byte, or leaving SP pointing at the stored data instead of the next free slot.

Source. Motorola M68HC11 Reference Manual (PUSH semantics, stack); Hamacher et al., §2.6 (byte order, stacks).

Question 5 of 6: Big-endian storage and the stack

Question 5: Big-endian storage and the stack (12 marks)