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9618Paper 1 · Theory Fundamentals§4.1, §4.2, §4.3

4. Processor Fundamentals

Von Neumann CPU, registers, buses, the FDE cycle with register transfer notation, interrupts, assembly language, addressing modes and bit manipulation.

Statometer76BankerNext Paper 189%
Marks a paper13.1 · 18%Rank#6 of 12 · #4 on P1Trend · last 12Oct/Nov 24 · 11: 20 marksOct/Nov 24 · 12: 23 marksOct/Nov 24 · 13: 6 marksMay/Jun 25 · 11: 21 marksMay/Jun 25 · 12: 17 marksMay/Jun 25 · 13: 19 marksOct/Nov 25 · 11: 11 marksOct/Nov 25 · 12: 10 marksOct/Nov 25 · 13: 11 marksMay/Jun 26 · 11: 9 marksMay/Jun 26 · 12: 0 marksMay/Jun 26 · 13: 9 marks
9 in 10 chance in the next paper

Everything for this topic — study hub

AS Level · 9618 · Paper 1

Statometer — what 33 real papers say about this topic and each of its 4 syllabus bullets

Banker · #6 of 12 in AS Level · recomputed with every new session

76BANKER
Banker#6 of 12 in AS Level#4 on Paper 1 Easing

Set in nearly every paper and worth a big slice of it — revise first, expect it.

Next Paper 1
89%
9 in 10 chance it is set
Marks a paper
13.1 / 75
18% of Paper 1 · fair share 13%
Appeared in
32 / 33
Paper 1 sittings 20212026
Last set
May/Jun 2026
9618/13 · Q7 · 9 marks · 11-series streak
Marks in each of the last 12 Paper 1 sittingsOct/Nov 24May/Jun 26
Oct/Nov 24 · 11: 20 marksOct/Nov 24 · 12: 23 marksOct/Nov 24 · 13: 6 marksMay/Jun 25 · 11: 21 marksMay/Jun 25 · 12: 17 marksMay/Jun 25 · 13: 19 marksOct/Nov 25 · 11: 11 marksOct/Nov 25 · 12: 10 marksOct/Nov 25 · 13: 11 marksMay/Jun 26 · 11: 9 marksMay/Jun 26 · 12: 0 marksMay/Jun 26 · 13: 9 marks

What the papers say

  • Set in 32 of 33 Paper 1 sittings on the current syllabus — treat it as certain.
  • Worth about 13.1 marks a paper (18% of Paper 1, 1.4× its fair share).
  • Last set May/Jun 2026 · 9618/13 · Q7 for 9 marks — in the most recent series.
  • Set in each of the last 11 series without a miss.
  • Easing off: about 16.3 marks a paper earlier, 10.8 in the latest years — still examined, just smaller.
  • Lives on “Complete” and “Identify” — 79% of its questions: you must produce something — code, a diagram, a table — practise doing it, not reading it.
  • 65% of its questions involve a diagram, table or figure — practise with pen and paper.
  • 67% of its questions are set out as code, pseudocode or a table to complete.
  • Its biggest question so far: 17 marks (May/Jun 2021 · 9618/12 · Q5).
  • Inside the topic, §4.1 CPU architecture carries the most marks (50%) and §4.3 Bit manipulation the least (3%).
  • It is examined mostly as AO1 (Knowledge & understanding, 57%), the rest AO2 (43%) — definitions and descriptions in syllabus words score.
  • The examiner has commented on 31 of its questions — read “What the examiner said” before you practise.

Command words

Share of questions using the word (a question can use several). What each wants →

Question shapes

  • ≤ 6 mk5
  • 7–9 mk19
  • 10–12 mk15
  • 13–15 mk7
  • 16+ mk2

Average 9.8 marks a question · 65% with a figure or table · 67% with code · biggest 17 marks

Assessment objectives — how it is examined

Every part of every current-syllabus question filed under Cambridge's AO1 / AO2 / AO3 (from its command word and what it asks you to do), so you know whether this topic pays for definitions, for applying, or for judging and building.

  • AO1 Knowledge & understanding
  • AO2 Apply & analyse
  • AO3 Design, program & evaluate

Paper 1 as a whole

Paper 1SyllabusMeasured
AO1 Knowledge & understanding60%68%
AO2 Apply & analyse40%32%
AO3 Design, program & evaluate0%0%

Syllabus = Cambridge's grid; measured = the bank's current-syllabus papers.

Inside the topic — every syllabus bullet, measured

Each part of each question is filed under the bullet it examines; the numbers are per Paper 1 sitting, exactly like the topic's. Open a bullet for its own Statometer.

  • 4.1CPU architecture#4 of 25 on Paper 1Banker · 8580% next paper5.7 marks29/33 sittings May/Jun 2026

    Asked in nearly every paper — the bullet to know cold.Syllabus: Von Neumann and the stored program concept; registers (PC, MDR, MAR, ACC, IX, CIR, status); ALU, CU, clock, IAS; the three buses; performance factors; USB, HDMI and VGA ports

    Next Paper 1
    80%
    8 in 10
    Marks a paper
    5.7
    8% of the paper · 50% of the topic
    Asked in
    29 / 33
    Paper 1 sittings · 42 questions
    Last asked
    May/Jun 2026
    9618/11 · Q3 · 2 marks · 11-series streak
    • Asked in 29 of 33 Paper 1 sittings — nearly every paper.
    • About 5.7 marks a paper (8% of Paper 1; 50% of the topic's marks across its 4 bullets).
    • Last asked May/Jun 2026 · 9618/11 · Q3 (2 marks) — in the most recent series.
    • Asked in each of the last 11 series.
    • Easing: 7.5 → 4.3 marks a paper.
    • Usually “Complete” or “Describe”: you must produce something — code, a diagram, a table — practise doing it, not reading it.
    • Biggest chunk of marks so far: 12 in Oct/Nov 2023 · 9618/11 · Q5.
    • It is examined mostly as AO1 (Knowledge & understanding, 66%), the rest AO2 (34%) — definitions and descriptions in syllabus words score.
    Last 12 sittings Easing
    Oct/Nov 24 · 11: 11 marksOct/Nov 24 · 12: 10.5 marksOct/Nov 24 · 13: 1.5 marksMay/Jun 25 · 11: 11 marksMay/Jun 25 · 12: 5 marksMay/Jun 25 · 13: 8 marksOct/Nov 25 · 11: 8.5 marksOct/Nov 25 · 12: 6 marksOct/Nov 25 · 13: 6 marksMay/Jun 26 · 11: 2 marksMay/Jun 26 · 12: 0 marksMay/Jun 26 · 13: 0 marks

    Assessment objectives

    • AO1 Knowledge & understanding
    • AO2 Apply & analyse
    • AO3 Design, program & evaluate
    • Complete52%
    • Describe48%
    • Identify45%
    • Explain36%
  • 4.1The fetch–execute cycle in register transfer notation; interrupts#23 of 25 on Paper 1Occasional · 2735% next paper1.3 marks14/33 sittings May/Jun 2025

    Rotated in occasionally — the bullet students skip and then meet.Syllabus: causes, applications, the ISR, and when and how interrupts are detected and handled

    Next Paper 1
    35%
    1 in 4
    Marks a paper
    1.3
    2% of the paper · 13% of the topic
    Asked in
    14 / 33
    Paper 1 sittings · 15 questions
    Last asked
    May/Jun 2025
    9618/13 · Q2 · 2 marks · 2 series ago
    • Asked in 14 of 33 Paper 1 sittings — roughly one paper in 2.
    • About 1.3 marks a paper (2% of Paper 1; 13% of the topic's marks across its 4 bullets).
    • Last asked May/Jun 2025 · 9618/13 · Q2 (2 marks), 2 seriess ago.
    • Easing: 2.4 → 0.6 marks a paper.
    • Usually “Describe” or “Identify”: full sentences with a reason, not one-word answers.
    • Biggest chunk of marks so far: 6.5 in May/Jun 2022 · 9618/13 · Q2.
    • It is examined almost entirely as AO1 (Knowledge & understanding, 81%) — definitions and descriptions in syllabus words score.
    Last 12 sittings Easing
    Oct/Nov 24 · 11: 0 marksOct/Nov 24 · 12: 4 marksOct/Nov 24 · 13: 0 marksMay/Jun 25 · 11: 0 marksMay/Jun 25 · 12: 5 marksMay/Jun 25 · 13: 2 marksOct/Nov 25 · 11: 0 marksOct/Nov 25 · 12: 0 marksOct/Nov 25 · 13: 0 marksMay/Jun 26 · 11: 0 marksMay/Jun 26 · 12: 0 marksMay/Jun 26 · 13: 0 marks

    Assessment objectives

    • AO1 Knowledge & understanding
    • AO2 Apply & analyse
    • AO3 Design, program & evaluate
    • Describe60%
    • Identify53%
    • Explain47%
    • Complete47%
  • 4.2Assembly language#7 of 25 on Paper 1Banker · 7680% next paper4.2 marks28/33 sittings May/Jun 2026

    Asked in nearly every paper — the bullet to know cold.Syllabus: machine code; the two-pass assembler; tracing programs; instruction groups (data movement, I/O, arithmetic, jumps, compare); the five addressing modes

    Next Paper 1
    80%
    8 in 10
    Marks a paper
    4.2
    6% of the paper · 34% of the topic
    Asked in
    28 / 33
    Paper 1 sittings · 28 questions
    Last asked
    May/Jun 2026
    9618/13 · Q7 · 4 marks · 11-series streak
    • Asked in 28 of 33 Paper 1 sittings — about 8 papers in 10.
    • About 4.2 marks a paper (6% of Paper 1; 34% of the topic's marks across its 4 bullets).
    • Last asked May/Jun 2026 · 9618/13 · Q7 (4 marks) — in the most recent series.
    • Asked in each of the last 11 series.
    • Usually “Complete” or “Write”: you must produce something — code, a diagram, a table — practise doing it, not reading it.
    • Biggest chunk of marks so far: 8 in May/Jun 2025 · 9618/13 · Q5.
    • It is examined mostly as AO2 (Apply & analyse, 73%), the rest AO1 (27%) — you must apply it to the given data or scenario — work it out, trace it, explain it in context.
    Last 12 sittings Steady
    Oct/Nov 24 · 11: 0 marksOct/Nov 24 · 12: 5.5 marksOct/Nov 24 · 13: 4.5 marksMay/Jun 25 · 11: 4.5 marksMay/Jun 25 · 12: 4.5 marksMay/Jun 25 · 13: 8 marksOct/Nov 25 · 11: 2 marksOct/Nov 25 · 12: 4 marksOct/Nov 25 · 13: 5 marksMay/Jun 26 · 11: 7 marksMay/Jun 26 · 12: 0 marksMay/Jun 26 · 13: 4 marks

    Assessment objectives

    • AO1 Knowledge & understanding
    • AO2 Apply & analyse
    • AO3 Design, program & evaluate
    • Complete64%
    • Write50%
    • Identify36%
    • Trace25%
  • 4.3Bit manipulation#25 of 25 on Paper 1Occasional · 2036% next paper0.6 marks9/33 sittings May/Jun 2026

    Rotated in occasionally — the bullet students skip and then meet.Syllabus: logical, arithmetic and cyclic shifts; AND, OR and XOR masks to test and set bits; monitoring and controlling devices with bit patterns

    Next Paper 1
    36%
    1 in 4
    Marks a paper
    0.6
    1% of the paper · 3% of the topic
    Asked in
    9 / 33
    Paper 1 sittings · 9 questions
    Last asked
    May/Jun 2026
    9618/13 · Q7 · 5 marks · 4-series streak
    • Asked in 9 of 33 Paper 1 sittings — roughly one paper in 4.
    • About 0.6 marks a paper (1% of Paper 1; 3% of the topic's marks across its 4 bullets).
    • Last asked May/Jun 2026 · 9618/13 · Q7 (5 marks) — in the most recent series.
    • Asked in each of the last 4 series.
    • Usually “Write” or “Complete”: you must produce something — code, a diagram, a table — practise doing it, not reading it.
    • Biggest chunk of marks so far: 5 in May/Jun 2026 · 9618/13 · Q7.
    • It is examined mostly as AO2 (Apply & analyse, 66%), the rest AO1 (35%) — you must apply it to the given data or scenario — work it out, trace it, explain it in context.
    Last 12 sittings Steady
    Oct/Nov 24 · 11: 0 marksOct/Nov 24 · 12: 3 marksOct/Nov 24 · 13: 0 marksMay/Jun 25 · 11: 0.5 marksMay/Jun 25 · 12: 0.5 marksMay/Jun 25 · 13: 1 marksOct/Nov 25 · 11: 0.5 marksOct/Nov 25 · 12: 0 marksOct/Nov 25 · 13: 0 marksMay/Jun 26 · 11: 0 marksMay/Jun 26 · 12: 0 marksMay/Jun 26 · 13: 5 marks

    Assessment objectives

    • AO1 Knowledge & understanding
    • AO2 Apply & analyse
    • AO3 Design, program & evaluate
    • Write67%
    • Complete44%
    • Identify44%
    • Trace33%

11% of the topic's marks sit in question parts that belong to another topic (scenario questions cross sections) or that no bullet claims; they count for the topic, not for a bullet.

Marks a paper, year by year

212223242526

By exam series

  • May/Jun17/18 · 12.4 mk
  • Oct/Nov15/15 · 16.5 mk

What you need to know4syllabus §4.1, §4.2, §4.3

  1. 4.1CPU architectureVon Neumann and the stored program concept; registers (PC, MDR, MAR, ACC, IX, CIR, status); ALU, CU, clock, IAS; the three buses; performance factors; USB, HDMI and VGA ports
  2. 4.1The fetch–execute cycle in register transfer notation; interruptscauses, applications, the ISR, and when and how interrupts are detected and handled
  3. 4.2Assembly languagemachine code; the two-pass assembler; tracing programs; instruction groups (data movement, I/O, arithmetic, jumps, compare); the five addressing modes
  4. 4.3Bit manipulationlogical, arithmetic and cyclic shifts; AND, OR and XOR masks to test and set bits; monitoring and controlling devices with bit patterns

Video lectures49ZAK's YouTube channel · play here

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Infographics5draw these the way the examiner expects · download as PNG

Two's complement, overflow & shifts8-bit two's complement: the MSB is worth −128. Range −128 … +127. Same bits, different meaning — saywhich you are using.Negating +37 → −37+370010010132 + 4 + 1invert11011010flip every bit (one’s complement)+111011011= −128 + 64 + 16 + 8 + 2 + 1 = −37 ✓Shortcut: from the right, copy up to and including the first 1, then flip the rest.Overflow0111111112700000001+ 110000000= −128 ✗ does not fit in 8 bitsOverflow = the answer needs more bits than the register has.Sign flips: two positives give a negative (or vice-versa).ShiftsShift0001 0110 (22) →logical left 10010 1100 (44)logical right 10000 1011 (11)arithmetic right 1keeps the sign bit1111 0100 (−12) →1111 1010 (−6)Left ×2, right ÷2. Bits shifted out are lost — that isanother cause of overflow. Logical fills with 0s.BCD: each denary digit as its own 4-bit nibble — 47 = 0100 0111. Used where exact decimals matter (currency).cswithzak.com

Two's complement, overflow & shifts

O LevelAS
Von Neumann architectureOne memory holds both instructions and data; CPU and memory talk over three buses.CPUControl Unitdecodes, sends signalsALUarithmetic & logicRegistersPC · MAR · MDR · CIR · ACCCachefast memory for frequently used dataMain memoryRAM — instructions + dataaddress bus →data bus ⇄control bus ⇄Address bus is unidirectional (CPU → memory). Its width sets the maximum addressable memory (n lines → 2ⁿ addresses).cswithzak.com

Von Neumann architecture

O LevelAS
Fetch – Decode – Execute cycleRegister transfer notation for the fetch stage — learn it word for word.FETCHinstruction from memoryDECODEcontrol unit interprets opcodeEXECUTEALU / registers / memoryrepeat (check for interrupts at the end of each cycle)1. MAR ← [PC]2. PC ← [PC] + 13. MDR ← [[MAR]]4. CIR ← [MDR]address of next instruction → MARincrement ready for the next fetchcontents of that address → MDRcopy the instruction into CIRcswithzak.com

Fetch–Decode–Execute cycle

O LevelAS
How an interrupt is handledA signal that tells the processor an event needs attention now — without it, the CPU would have to keeppolling every device.SourcesHardwarekey press / mouse clickprinter out of papertimer (time-slice ended)hardware fault, power failureSoftwaredivision by zeroarithmetic overflowillegal / undefined instructiontwo processes need one resource1Device or program raises an interrupt → interrupt flag is set2CPU finishes the current fetch–decode–execute cycle, then tests the flag3Higher priority? Push the registers (PC, ACC, …) onto the stack4Load the ISR (interrupt service routine) address into the PC; run it5Pop the saved registers off the stack; resume the interrupted programPriorities: a low-priority interrupt can itself be interrupted by a higher one (nested); lower ones wait in a queue.The interrupt flag is tested once per cycle — that is why “check for interrupts” is the last step of the FDE cycle.Buffers + interrupts let a slow printer work while the CPU carries on: it interrupts only when its buffer needs refilling.cswithzak.com

How an interrupt is handled

O LevelAS
Assembly language & addressing modesInstruction = opcode + operand. The addressing mode says how to interpret the operand. Trace it againstthe memory below.MemoryAddressContents2003002017202930042IX = 2 ACC = ?InstructionModeWhat the operand meansACC ←LDM #200immediatethe number itself200LDD 200directthe contents of address 200300LDI 200indirectaddress 200 holds the address of the value42LDX 200indexedaddress 200 + IX = 202, load its contents9LDR #2immediateload 2 into the index register IXInstruction set you must knowSTO <addr> store ACC ADD/SUB <addr> or #n INC/DEC <reg>CMP <addr> or #n compare with ACC CMI <addr> indirectJMP <addr> jump JPE / JPN <addr> jump if equal / notIN / OUT char ↔ ACC AND / OR / XOR LSL / LSR #n ENDBit manipulation with masksAND #B00001111 → clear the top 4 bits (test / isolate bits)OR #B10000000 → set bit 7 XOR #B11111111 → flip all bitsRelative: operand is an offset from the PC(jump forward/back n instructions).Operands: #n denary, #Bn binary, #&n hex.Labels stand for addresses; the assemblerresolves them via the symbol table.Absolute = a fixed address (direct).Register transfer: MAR ← [PC] · PC ← [PC] + 1 · MDR ← [[MAR]] · CIR ← [MDR]. Square brackets mean “contents of”.cswithzak.com

Assembly language & addressing modes

AS

Browse all infographics →

Key terms15use these exact words in the exam

registerMARMDRCIRPCACCRTNinterruptassemblyopcodeoperandaddressing modelogical shiftarithmetic shiftmask

Dotted terms are defined in the glossary.

Code help4referenced to the Cambridge pseudocode guide

Assembly: add two numbers and store

assembly Run in Assembler
LDD 200 ; load contents of address 200 into ACC (direct)
ADD 201 ; add contents of address 201
STO 202 ; store ACC at address 202
END

💡 Direct addressing: the operand is an address. LDM #5 would load the number 5 itself (immediate).

Assembly: loop with a counter and conditional jump

assembly Run in Assembler
LDM #0 ; ACC ← 0 (total)
STO 300
LDM #5 ; counter
STO 301
LOOP: LDD 300
ADD 301 ; total ← total + counter
STO 300
LDD 301
DEC ACC ; counter ← counter − 1
STO 301
CMP #0
JPN LOOP ; jump if not equal to 0
LDD 300
OUT
END

Bit masking to test a flag

assembly Run in Assembler
LDD 400 ; status byte
AND #B00000100 ; keep only bit 2
CMP #0
JPE OFF ; bit 2 was 0
; ... bit 2 set
OFF: END

Simulate shifts and masks in pseudocode

pseudocode Run in Playground
DECLARE ACC : INTEGER
ACC 45 // 0010 1101
OUTPUT ACC * 2 // LSL #1 → 90
OUTPUT ACC DIV 4 // LSR #2 → 11
OUTPUT ACC MOD 16 // AND #B00001111 → 13 (low nibble)

Playground examples2runnable programs for this topic

  • Bit manipulation — test, set and clear a bit

    DIV and MOD by powers of 2 do the job of shifts and masks (AND/OR/XOR) from the 9618 instruction set.

    ASPseudocodeTheory in code 9618 §4.4
    Run
  • Fetch–decode–execute simulator

    A tiny CPU: PC and ACC registers, instructions in memory, one loop that fetches, decodes with CASE and executes.

    ASPseudocodeTheory in code 9618 §4.1, §4.3
    Run

Assembler34programs to step through the CPU simulator

  • Add two numbers in memory

    The classic first program: load, add, store. Watch ACC change and address 202 get written.

    ASLoading & storing 9618 §4.2
    Run
  • LDM #n vs LDD n — number or address?

    The most-tested distinction: LDM #200 loads the number 200, LDD 200 loads the CONTENTS of address 200.

    ASLoading & storing 9618 §4.2
    Run
  • Symbolic addresses (labels) instead of numbers

    Label your data and jump targets — the assembler turns each label into an address. Check the memory panel to see where NUM1 landed.

    ASLoading & storing 9618 §4.2, §5.2 assembler
    Run
  • LDR #n and MOV IX — filling the index register

    Two ways to set IX: load a number straight into it (LDR) or copy ACC across (MOV).

    ASLoading & storing 9618 §4.2
    Run
  • Swap two memory locations

    No swap instruction exists — do it through ACC with a temporary location, like Temp ← X in pseudocode.

    ASLoading & storing 9618 §4.2
    Run
  • ADD and SUB with immediate values

    ADD #n adds the number n; ADD n adds the contents of address n. Both forms are in the syllabus.

    ASArithmetic 9618 §4.2
    Run
  • INC and DEC on ACC and IX

    Both registers can be incremented and decremented directly — the usual way to move a counter or an index.

    ASArithmetic 9618 §4.2
    Run
  • Multiply by repeated addition

    There is no MUL — 6 × 4 is 4 added six times, counted down with DEC and CMP/JPN.

    ASA2Arithmetic 9618 §4.2
    Run
  • Negative results and two's complement

    5 − 12 = −7. Look at the binary view of ACC: 11111001 is −7 in 8-bit two's complement.

    ASArithmetic 9618 §1.1, §4.2
    Run
  • Overflow — when the result no longer fits

    In an 8-bit accumulator 200 + 100 = 300 cannot be represented. The simulator keeps the value but raises the overflow flag, exactly what the exam wants you to explain.

    ASArithmetic 9618 §1.1 overflow
    Run
  • Immediate addressing

    The operand IS the value. LDM #25 puts 25 in ACC — no memory read happens in the execute stage.

    ASA2Addressing modes 9618 §4.2, §20.1
    Run
  • Direct addressing

    The operand is the address of the value. LDD 200 reads memory location 200. Watch MAR/MDR in the execute stage.

    ASA2Addressing modes 9618 §4.2, §20.1
    Run
  • Indirect addressing (LDI)

    The operand holds the address of the address. [200] = 300, so LDI 200 loads [300]. A pointer in assembly.

    ASA2Addressing modes 9618 §4.2, §20.1
    Run
  • Indexed addressing (LDX) — walking an array

    Address = operand + IX. With IX = 0, 1, 2… LDX 300 visits 300, 301, 302 — how a loop reads an array.

    ASA2Addressing modes 9618 §4.2, §20.1
    Run
  • All five addressing modes in one program

    Immediate, direct, indirect, indexed and relative, one after another — the summary table you should be able to reproduce in Paper 3.

    A2Addressing modes 9618 §4.2, §20.1
    Run
  • IF … THEN … ELSE with CMP and JPE

    Compare sets a flag; JPE / JPN choose the path. This is how selection is built in a low-level language.

    ASCompare & jump 9618 §4.2
    Run
  • Counted loop: total 1 to 5

    A FOR loop in assembly — counter in memory, CMP against the limit, JPN back to the top.

    ASCompare & jump 9618 §4.2
    Run
  • Sum an array — the working idiom

    Keep the loop counter in memory alongside IX: the counter is compared, IX does the indexing. Ends when COUNT = 0.

    ASA2Compare & jump 9618 §4.2
    Run
  • Linear search through memory

    Search for 30 in an array of 5 using LDX; output Y if found, N if not. Change TARGET to try both paths.

    ASA2Compare & jump 9618 §4.2
    Run
  • CMI — compare through a pointer

    CMI 200 compares ACC with the value at the address stored in 200 — indirect addressing for a compare.

    ASA2Compare & jump 9618 §4.2
    Run
  • AND mask — test whether a bit is set

    AND with B00000100 keeps only bit 2. If the result is 0 the bit was clear. Watch the binary working in the execute panel.

    ASBit manipulation 9618 §4.3
    Run
  • OR — set a bit without touching the others

    OR with B00010000 forces bit 4 to 1 and leaves the rest alone.

    ASBit manipulation 9618 §4.3
    Run
  • XOR — toggle bits and clear ACC

    XOR flips every bit where the mask has a 1; XOR-ing a value with itself gives 0 (a classic trick).

    ASBit manipulation 9618 §4.3
    Run
  • LSL / LSR — multiply and divide by powers of 2

    Shifting left 1 doubles, right 1 halves (integer). 6 ≪ 2 = 24, 24 ≫ 3 = 3.

    ASBit manipulation 9618 §4.3
    Run
  • Bits lost off the end of a logical shift

    B11000011 shifted left 2 loses its two top 1s: the result is B00001100. Logical shifts never bring them back.

    ASBit manipulation 9618 §4.3
    Run
  • Clear one bit with AND and a mask

    To clear bit 0 AND with B11111110 — every bit kept except the one under the 0.

    ASBit manipulation 9618 §4.3
    Run
  • IN and OUT — echo a character

    IN reads one keyboard character into ACC as its ASCII code; OUT prints the character whose code is in ACC.

    ASInput & output 9618 §4.2
    Run
  • Lower-case to upper-case with SUB #32

    ASCII 'a' is 97 and 'A' is 65 — the difference is 32, so subtract it (or AND with B11011111).

    ASInput & output 9618 §4.2, §1.2 ASCII
    Run
  • Print a string stored in memory

    Characters live at consecutive addresses, ending with 0. LDX + INC IX walks them until the terminator.

    ASA2Input & output 9618 §4.2
    Run
  • Count how many characters were typed

    Read characters until a full stop; count them and output the count as a digit (count + 48 = its ASCII).

    ASA2Input & output 9618 §4.2
    Run
  • Trace question: what is in 320 at the end?

    Typical Paper 1 question. Trace ACC and the memory locations, then check against the simulator's trace table.

    ASExam-style traces 9618 §4.2
    Run
  • Trace question: loop with JPN

    Complete the trace table for ACC and address 500 — how many times does the loop body run?

    ASExam-style traces 9618 §4.2
    Run
  • Trace question: masks and shifts

    Paper 1 bit-manipulation trace: show ACC in binary after each instruction.

    ASExam-style traces 9618 §4.3
    Run
  • Trace question: indirect and indexed together (A2)

    Paper 3 style: combine LDI, LDX and CMI and give the final contents of the registers.

    A2Exam-style traces 9618 §20.1
    Run

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