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Byte Memory

Gather four kinds of evidence: prediction of relevant state; behavior across varied cases including boundaries; explanation using trace/state evidence; and a reasoned correction of one mismatch. Mark each as emerging, with support or independent using paper or your school’s existing approved process. A challenge pass is machine-behavior evidence, not proof of authorship or complete mastery. Accept equivalent correct programs; the public key is a model, not the only possible answer.

Public teaching material. Solutions are examples; equivalent correct programs may also pass the actual checks.

7. Memory: The Computer’s Notebook

Store and actually retrieve a byte after overwriting R0.

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1. Predict and trace

Before running the Try program, predict its output and trace the first three executed instructions. Track the relevant registers, flags or memory as needed. Then step the program to compare.

LOAD 42
STORE 0
LOAD 99
STORE 1
LOAD 0
LOAD [0]
PRINT
LOAD [1]
PRINT
HALT

Queue input: None

Initialize memory: All bytes initially zero

Try prediction key

Output: 42, 99

InstructionPC beforePC afterR0 beforeR0 afterRelevant stateOutput
LOAD01042{"registers":[42,0,0,0,0,0,0,0],"flags":{"zero":false,"negative":false,"overflow":false},"sp":255,"stack":[],"callStack":[],"loopStack":[],"memoryReads":[],"memoryWrites":[]}None
STORE124242{"registers":[42,0,0,0,0,0,0,0],"flags":{"zero":false,"negative":false,"overflow":false},"sp":255,"stack":[],"callStack":[],"loopStack":[],"memoryReads":[],"memoryWrites":[0]}None
LOAD234299{"registers":[99,0,0,0,0,0,0,0],"flags":{"zero":false,"negative":false,"overflow":false},"sp":255,"stack":[],"callStack":[],"loopStack":[],"memoryReads":[],"memoryWrites":[]}None

2. Build and check

Read a byte-valued input (0–255), save it at memory address 10, clear R0, then retrieve it from memory and print it.

Required instruction types: INPUT, STORE, LOAD

Example challenge solution

INPUT
STORE 10
LOAD 0
LOAD [10]
PRINT
HALT

Actual checker fixtures

Case 1
Input
9
Initial memory
All bytes initially zero
Expected output
9
Expected final memory
10: 9
Required memory reads
10
Case 2
Input
42
Initial memory
All bytes initially zero
Expected output
42
Expected final memory
10: 42
Required memory reads
10
Case 3
Input
127
Initial memory
All bytes initially zero
Expected output
127
Expected final memory
10: 127
Required memory reads
10

3. Explain the machine

Which code reads memory 10, and which loads the number10?

Reasoning and teaching note

LOAD [10] reads the byte; LOAD 10 is literal. STORE and later retrieval preserve the byte beyond R0 overwrite.

8. Addresses in Registers

Use an address as data and read a lookup table.

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1. Predict and trace

Before running the Try program, predict its output and trace the first three executed instructions. Track the relevant registers, flags or memory as needed. Then step the program to compare.

STORE 12 0
STORE 99 7
LOAD 7 R1
LOAD [R1]
PRINT
HALT

Queue input: None

Initialize memory: 0: 12

Try prediction key

Output: 99

InstructionPC beforePC afterR0 beforeR0 afterRelevant stateOutput
STORE0100{"registers":[0,0,0,0,0,0,0,0],"flags":{"zero":false,"negative":false,"overflow":false},"sp":255,"stack":[],"callStack":[],"loopStack":[],"memoryReads":[],"memoryWrites":[0]}None
STORE1200{"registers":[0,0,0,0,0,0,0,0],"flags":{"zero":false,"negative":false,"overflow":false},"sp":255,"stack":[],"callStack":[],"loopStack":[],"memoryReads":[],"memoryWrites":[7]}None
LOAD2300{"registers":[0,7,0,0,0,0,0,0],"flags":{"zero":false,"negative":false,"overflow":false},"sp":255,"stack":[],"callStack":[],"loopStack":[],"memoryReads":[],"memoryWrites":[]}None

2. Build and check

Read an address from INPUT, copy it to R1, retrieve the seeded byte at that address, and print the byte. The checker changes both address and stored value.

Required instruction types: INPUT, COPY, LOAD

Example challenge solution

INPUT
COPY R0 R1
LOAD [R1]
PRINT
HALT

Actual checker fixtures

Case 1
Input
0
Initial memory
0: 12
Expected output
12
Required memory reads
0
Case 2
Input
7
Initial memory
7: 99
Expected output
99
Required memory reads
7
Case 3
Input
255
Initial memory
255: 88
Expected output
88
Required memory reads
255

3. Explain the machine

How do LOAD R1 and LOAD [R1] differ when R1=7 and memory 7=99?

Reasoning and teaching note

First reads7 from the register; second reads99 from address 7.

9. Bytes and Boundaries

See why memory wraps while registers do not.

Open this lesson →

1. Predict and trace

Before running the Try program, predict its output and trace the first three executed instructions. Track the relevant registers, flags or memory as needed. Then step the program to compare.

LOAD 256
PRINT
STORE 255
LOAD [255]
PRINT
LOAD -1
STORE 255
LOAD [255]
PRINT
HALT

Queue input: None

Initialize memory: All bytes initially zero

Try prediction key

Output: 256, 0, 255

InstructionPC beforePC afterR0 beforeR0 afterRelevant stateOutput
LOAD010256{"registers":[256,0,0,0,0,0,0,0],"flags":{"zero":false,"negative":false,"overflow":false},"sp":255,"stack":[],"callStack":[],"loopStack":[],"memoryReads":[],"memoryWrites":[]}None
PRINT12256256{"registers":[256,0,0,0,0,0,0,0],"flags":{"zero":false,"negative":false,"overflow":false},"sp":255,"stack":[],"callStack":[],"loopStack":[],"memoryReads":[],"memoryWrites":[]}256
STORE23256256{"registers":[256,0,0,0,0,0,0,0],"flags":{"zero":false,"negative":false,"overflow":false},"sp":255,"stack":[],"callStack":[],"loopStack":[],"memoryReads":[],"memoryWrites":[255]}256

2. Build and check

Read an integer, store it at address 255, retrieve the resulting byte, and print it. Do not calculate a substitute fixed answer.

Required instruction types: INPUT, STORE, LOAD

Example challenge solution

INPUT
STORE 255
LOAD [255]
PRINT
HALT

Actual checker fixtures

Case 1
Input
255
Initial memory
All bytes initially zero
Expected output
255
Expected final memory
255: 255
Required memory reads
255
Case 2
Input
256
Initial memory
All bytes initially zero
Expected output
0
Expected final memory
255: 0
Required memory reads
255
Case 3
Input
-1
Initial memory
All bytes initially zero
Expected output
255
Expected final memory
255: 255
Required memory reads
255
Case 4
Input
511
Initial memory
All bytes initially zero
Expected output
255
Expected final memory
255: 255
Required memory reads
255

3. Explain the machine

Why is value256 storable at address 255, while address 256 is invalid?

Reasoning and teaching note

Byte data wraps256→0; addresses are independently bounded0..255 and never wrap.