Level:
Correct: 0
Incorrect: 0
Marks: 0/0
Exam Practice
Subject:
Topics:
Cambridge IGCSE Computer Science (Paper 1) Smart practice A mixed round aimed at your weak spots
Unit 1
Data representation
Number systems
Binary, denary and hexadecimal conversions
0%
Uses of hexadecimal
0%
Binary addition and overflow
0%
Logical binary shifts
0%
Two's complement
0%
Text, sound and images
Character sets: ASCII and Unicode
0%
Representing sound
0%
Representing images
0%
Data storage and compression
Data storage units and file size calculations
0%
Data compression
0%
Unit 2
Data transmission
Types and methods of data transmission
Packets and packet switching
0%
Methods of data transmission
0%
The USB interface
0%
Methods of error detection
Parity, checksum and echo check
0%
Check digits and ARQ
0%
Encryption
Encryption
0%
Unit 3
Hardware
Computer architecture
The CPU, Von Neumann architecture and the FDE cycle
0%
Cores, cache and clock speed
0%
Instruction sets and embedded systems
0%
Input and output devices
Input devices
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Output devices
0%
Sensors
0%
Data storage
Primary storage: RAM and ROM
0%
Secondary storage: magnetic, optical and solid-state
0%
Virtual memory and cloud storage
0%
Network hardware
Network hardware: NIC, MAC and IP addresses, routers
0%
Unit 4
Software
Types of software and interrupts
System software, application software and operating systems
0%
Firmware and interrupts
0%
Programming languages, translators and IDEs
High-level languages, low-level languages and assembly language
0%
Compilers and interpreters
0%
Integrated development environments (IDEs)
0%
Unit 5
The internet and its uses
The internet and the world wide web
The internet, the world wide web, URLs and HTTP(S)
0%
Web browsers, locating web pages and cookies
0%
Digital currency
Digital currency and blockchain
0%
Cyber security
Cyber security threats
0%
Keeping data safe: security solutions
0%
Unit 6
Automated and emerging technologies
Automated systems
Automated systems
0%
Robotics
Robotics
0%
Artificial intelligence
Artificial intelligence
0%
Avg
score:
–
Topics
done:
0/39

High-level languages, low-level languages and assembly language

Practise 26 exam-style questions on High-level languages, low-level languages and assembly language for Cambridge IGCSE Computer Science (Paper 1). Write your answer as you would in the exam and get it marked in seconds against the mark scheme, point by point, with an AI tutor on hand if you get stuck.

  • 26 questions
  • 1–6 marks each
  • Instant marking
  • No sign-up needed
20 questions, marked as you go

Sample questions with model answers

Here are 4 of the questions in this topic, each with a model answer, an explanation of why it earns the marks and the mark scheme an examiner would use.

Question 1

1 mark

State the name of the only type of language that a processor can execute directly, without any translation.

Model answer

Machine code.

Explanation

The processor only understands machine code: instructions in binary. Every other language, including assembly language, must be translated into machine code before it can be executed.

Mark scheme
  1. 1 mark

    Machine code

    accept machine language / binary instructions; do not accept assembly language or low-level language on its own

Question 2

2 marks

Give two advantages of writing a program in a high-level language rather than a low-level language.

Model answer

A high-level language is easier to read and write because it uses English-like statements. It is also machine independent, so the program can run on different types of processor.

Explanation

The syllabus comparison points are ease of reading and writing, ease of debugging, machine independence and direct manipulation of hardware. The first three favour high-level languages; only the last favours low-level languages.

Mark scheme
  1. 1 mark

    It is easier to read and write / understand (English-like)

    max 2

  2. 1 mark

    It is easier to debug

  3. 1 mark

    It is machine independent / portable

  4. 1 mark

    It is quicker to write / needs fewer lines of code

  5. 1 mark

    The programmer does not need to know about the processor / hardware

  6. 1 mark

    It is easier to maintain / update

    do not accept 'easier' on its own

Question 3

3 marks

Explain why a program written in assembly language for one type of processor may not run on a different type of processor.

Model answer

Assembly language is machine dependent. Each type of processor has its own set of instructions, so the machine code produced from the program will not be understood by a different processor. The program would have to be rewritten for the new processor.

Explanation

Because each assembly instruction maps to one machine code instruction of a particular processor, the program is tied to that processor's instruction set. High-level programs avoid this because they can be translated separately for each processor.

Mark scheme
  1. 1 mark

    Assembly language is machine dependent / specific to one type of processor

    max 3

  2. 1 mark

    Each type of processor has its own instruction set

  3. 1 mark

    The machine code produced will not be understood by the other processor

  4. 1 mark

    The program would have to be rewritten (and assembled again) for the other processor

Question 4

4 marks

The program is written in assembly language.

LDA 40
SUB 41
STO 42

Describe how this program is translated so that the processor can execute it.

Model answer

The program is translated by an assembler. The assembler converts each mnemonic, such as LDA, into its binary machine code, and converts the memory addresses 40, 41 and 42 into binary. Each of the three assembly instructions becomes exactly one machine code instruction, producing a machine code program that the processor can execute.

Explanation

Assembly language translation is simple compared with compiling: there is a one-to-one match between assembly instructions and machine code instructions, so the assembler essentially swaps each mnemonic and address for its binary equivalent.

Mark scheme
  1. 1 mark

    An assembler is used

    max 4

  2. 1 mark

    Each mnemonic (LDA, SUB, STO) is converted into its binary machine code (opcode)

  3. 1 mark

    The memory addresses (40, 41, 42) are converted into binary

  4. 1 mark

    Each assembly instruction becomes one machine code instruction

    accept 'one-to-one'

  5. 1 mark

    The result is a machine code program that the processor can execute

  6. 1 mark

    The assembler reports errors, e.g. an invalid mnemonic

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