📚 Year 10 Eduqas Computer Science: Core Knowledge Revision | Year 10 Eduqas 计算机：核心知识点梳理
This article distills the essential topics taught in Year 10 of the Eduqas GCSE Computer Science specification. You will revisit the architecture of the CPU, memory and storage, data representation, binary arithmetic, Boolean logic, networking, cybersecurity threats and fundamental algorithms. Each section pairs clear English explanations with Mandarin translations to reinforce understanding and support independent revision.
本文梳理了Eduqas GCSE 计算机科学 Year 10 的核心知识点，涵盖 CPU 架构、存储器、数据表示、二进制运算、布尔逻辑、网络、网络安全威胁以及基础算法。每个小节均配有中英双语解析，帮助你巩固理解并高效备考。
1. The CPU and Von Neumann Architecture | CPU 与冯·诺依曼架构
The Central Processing Unit (CPU) is the 'brain' of the computer, responsible for executing instructions stored in main memory. Eduqas expects you to describe the Von Neumann architecture, in which program instructions and data share the same memory space and are fetched sequentially over a system bus.
中央处理器（CPU）是计算机的“大脑”，负责执行存储在主存中的指令。Eduqas 要求你能够描述冯·诺依曼架构，即程序指令与数据共享同一内存空间，并通过系统总线按顺序被读取。
Key components include the Control Unit (CU), which decodes instructions and directs data flow; the Arithmetic Logic Unit (ALU), which performs calculations and logical operations; and registers such as the Program Counter (PC), Memory Address Register (MAR), Memory Data Register (MDR) and Accumulator (ACC). The fetch‑decode‑execute cycle repeats continuously: the address in the PC is placed on the address bus, the instruction is fetched into the MDR, the CU decodes it, and the ALU executes it, after which the PC increments to point to the next instruction.
关键部件包括控制单元（CU），负责译码指令并控制数据流动；算术逻辑单元（ALU），执行计算与逻辑运算；以及多个寄存器，如程序计数器（PC）、内存地址寄存器（MAR）、内存数据寄存器（MDR）和累加器（ACC）。取指‑译码‑执行周期不断循环：PC 中的地址被放到地址总线上，指令被取出送入 MDR，CU 译码，ALU 执行，随后 PC 递增指向下一条指令。
2. Main Memory (RAM & ROM) | 主存储器（RAM 与 ROM）
Random Access Memory (RAM) is volatile, meaning it loses its contents when the power is turned off. It stores the operating system, currently running applications and data in active use. Read Only Memory (ROM) is non‑volatile and retains its contents without power; it typically holds the boot firmware (BIOS) that loads when the computer is switched on.
随机存取存储器（RAM）是易失性的，掉电后数据丢失。它存放操作系统、正在运行的程序和正在使用的数据。只读存储器（ROM）是非易失性的，断电后仍能保存内容；通常用于存放启动固件（BIOS），电脑开机时加载。
Eduqas students should also understand that virtual memory uses a portion of the hard disk or SSD as an extension of RAM when physical RAM is full, though this is much slower and can cause disk thrashing.
学生还需理解虚拟内存：当物理 RAM 已满时，系统使用硬盘或固态硬盘的一部分作为 RAM 的扩展，但这会显著降低速度并可能引起磁盘颠簸。
3. Secondary Storage | 辅助存储器
Secondary storage provides long‑term, non‑volatile retention of data. The three main categories are magnetic (hard disk drives), optical (CDs, DVDs) and solid‑state (SSDs, USB flash drives). You must be able to compare them in terms of capacity, speed, portability, durability and cost per gigabyte.
辅助存储器提供长期、非易失的数据保存。三大类别为磁性（机械硬盘）、光学（CD、DVD）和固态（SSD、U盘）。你需要从容量、速度、便携性、耐用性和每 GB 成本等方面对它们进行比较。
For example, magnetic storage offers high capacity at low cost but contains moving parts prone to damage. Optical storage is cheap and portable but has limited capacity. Solid‑state storage is fast, shock‑resistant and energy‑efficient but typically more expensive per GB.
例如，磁性存储容量大、成本低，但因含机械部件而容易损坏；光学存储廉价便携但容量有限；固态存储速度快、抗震且节能，但每 GB 成本通常较高。
4. Data Representation: Binary and Hexadecimal | 数据表示：二进制与十六进制
Computers use binary (base‑2) because switches inside logic circuits represent only two states: 0 and 1. In the Eduqas specification, you need to convert unsigned integers between denary, binary and hexadecimal (base‑16), and understand why hexadecimal is used as a shorthand for long binary strings. Each hex digit corresponds to a nibble (4 bits).
计算机使用二进制（基数为 2）是因为逻辑电路中的开关仅表示两种状态：0 和 1。根据 Eduqas 大纲，你需要掌握无符号整数在十进制、二进制和十六进制（基数为 16）之间的转换，并理解为何十六进制被用作长二进制串的简写。每个十六进制位恰好对应一个半字节（4 位）。
For example, the 8‑bit binary number 1010 1100₂ can be split into two nibbles: 1010₂ (A in hex) and 1100₂ (C in hex), giving the hex value AC₁₆. Converting from denary to binary or hex requires repeated division by 2 or 16.
例如，8 位二进制数 1010 1100₂ 可以分成两个半字节：1010₂（十六进制 A）和 1100₂（十六进制 C），得到十六进制值 AC₁₆。从十进制转换到二进制或十六进制则需要连续除以 2 或 16 取余。
5. Binary Arithmetic and Units | 二进制运算与单位
ds the bit width of the register, an overflow error occurs, which can produce incorrect results. You should know the names and ratios of data units: bit, nibble (4 bits), byte (8 bits), kilobyte (1,000 bytes or 10³ bytes), megabyte (10⁶), gigabyte (10⁹) and terabyte (10¹²).
参加 GCSE 考试的学生必须掌握二进制加法规则：0+0 = 0，0+1 = 1，1+0 = 1，1+1 = 0 进 1，1+1+1 = 1 进 1。当计算结果超出寄存器的位宽时，会发生溢出错误，导致结果不正确。你还需要了解数据单位及其比例：位（bit）、半字节（nibble，4 位）、字节（byte，8 位）、千字节（kB，10³ 字节）、兆字节（MB，10⁶）、吉字节（GB，10⁹）和太字节（TB，10¹²）。
Binary shifts are also examined: shifting all bits left by one place multiplies the number by 2; shifting right by one place divides by 2 (integer division, discarding remainders). This is an efficient alternative to multiplication and division in low‑level programming.
二进制移位也是考点：将所有位左移一位相当于乘以 2，右移一位相当于除以 2（整除，丢弃余数）。在低级编程中，这可以高效地替代乘法和除法运算。
6. Logic Gates and Truth Tables | 逻辑门与真值表
You must be able to identify and use the AND, OR and NOT gates, and draw their truth tables. An AND gate outputs 1 only when both inputs are 1. An OR gate outputs 1 when at least one input is 1. A NOT gate (inverter) outputs the opposite of its single input.
你必须能够识别并运用与门（AND）、或门（OR）和非门（NOT），并绘制它们的真值表。与门仅在两个输入均为 1 时输出 1；或门在至少一个输入为 1 时输出 1；非门（反相器）则输出与其单一输入相反的值。
AND B) OR (NOT C). Eduqas questions often ask you to complete a truth table for a circuit containing two or more gates.
简单的逻辑电路可用布尔表达式描述，例如 Q = (A AND B) OR (NOT C)。Eduqas 试题常要求你为包含两个或多个门路的电路补全真值表。
| A | B | A AND B | A OR B | NOT A |
|---|---|---|---|---|
| 0 | 0 | 0 | 0 | 1 |
| 0 | 1 | 0 | 1 | 1 |
| 1 | 0 | 0 | 1 | 0 |
| 1 | 1 | 1 | 1 | 0 |
7. Computer Networks and Topologies | 计算机网络与拓扑结构
A network connects two or more computing devices so they can share resources such as files, printers and internet connections. Lan (Local Area Network) covers a small geographical area; Wan (Wide Area Network) spans large distances, often using third‑party communication links.
网络将两台或以上计算机连接起来，使之能够共享文件、打印机和互联网连接等资源。局域网（LAN）覆盖较小地理范围；广域网（WAN）跨越远距离，通常借助第三方通信链路。
Common topologies include the star, where all nodes connect to a central switch or hub, and the bus, where all nodes share a single backbone cable. The star topology offers easy fault isolation — if one cable fails, the rest of the network remains unaffected — while the bus topology is cheaper but suffers from single‑point‑of‑failure and signal collisions.
常见拓扑结构包括星型（所有节点连接到中央交换机或集线器）和总线型（所有节点共享一条主干电缆）。星型拓扑便于隔离故障——一根线缆损坏不影响其余部分；总线拓扑成本较低，但存在单点故障和信号冲突的问题。
You should also be able to discuss factors affecting network performance: bandwidth, number of users, transmission media (wireless, fibre optic, copper) and error rate.
你还需讨论影响网络性能的因素：带宽、用户数量、传输介质（无线、光纤、铜缆）以及误码率。
8. Network Protocols and Layers | 网络协议与分层
A protocol is a set of rules defining how devices communicate. The TCP/IP model organises network functions into four layers: Application (e.g. HTTP, FTP, SMTP), Transport (TCP, UDP), Internet (IP) and Link (Ethernet, Wi‑Fi). This layering simplifies troubleshooting and development because each layer works independently.
协议是一套规定设备如何通信的规则。TCP/IP 模型将网络功能分为四层：应用层（如 HTTP、FTP、SMTP）、传输层（TCP、UDP）、互联网层（IP）和链路层（以太网、Wi‑Fi）。分层使各层独立工作，从而简化了故障排查与开发。
Key protocols to remember: HTTP transmits web pages, HTTPS encrypts the data using SSL/TLS, FTP transfers files, SMTP sends emails, and POP3/IMAP retrieve emails. Eduqas may ask you to suggest which protocol to use in a given scenario.
需记住的关键协议：HTTP 传输网页，HTTPS 使用 SSL/TLS 加密数据，FTP 传输文件，SMTP 发送邮件，POP3/IMAP 收取邮件。Eduqas 可能会让你在特定场景中选用合适的协议。
9. Cybersecurity Threats and Prevention | 网络安全威胁与防护
Malware (malicious software) includes viruses, worms, trojans and ransomware. Social engineering tricks users into revealing confidential information; phishing emails pretend to be from trusted sources. Brute‑force attacks attempt to guess passwords by trying all possible combinations, while Denial‑of‑Service (DoS) attacks flood a server with traffic to make it unavailable.
恶意软件（Malware）包括病毒、蠕虫、木马和勒索软件。社会工程学通过欺骗用户获取机密信息；网络钓鱼邮件伪装成可信来源。暴力攻击尝试所有可能的密码组合以破解口令，而拒绝服务攻击（DoS）则通过流量洪泛使服务器无法提供服务。
Prevention strategies include using firewalls to filter traffic, encryption to protect data in transit, strong password policies, regular software updates (patching) and anti‑malware software. Penetration testing and ethical hacking help organisations identify vulnerabilities before criminals do.
防护策略包括：使用防火墙过滤流量、加密保护传输中的数据、推行强密码策略、定期软件更新（打补丁）以及防恶意软件。渗透测试与道德黑客可帮助机构在犯罪分子发现前找出漏洞。
10. Algorithms: Searching and Sorting | 算法：搜索与排序
An algorithm is a step‑by‑step procedure for solving a problem. Two search algorithms are required: linear search (checking each item in turn) and binary search (repeatedly dividing a sorted list in half). Binary search is much faster — its time complexity is logarithmic — but only works on sorted data.
算法是解决问题的分步过程。需要掌握两种搜索算法：线性搜索（逐一检查每个元素）和二分搜索（不断将有序列表对半分）。二分搜索要快得多，时间复杂度为对数级，但仅适用于已排序的数据。
For sorting, Eduqas focuses on bubble sort (repeatedly comparing adjacent items and swapping them if out of order) and merge sort (recursively splitting the list into single items and then merging sorted sub‑lists). You should be able to trace these algorithms step by step and understand their impact on performance.
排序方面，Eduqas 重点考查冒泡排序（反复比较相邻元素，若顺序错误则交换）和归并排序（递归地拆分列表至单个元素，再合并已排序的子列表）。你需要逐步跟踪这些算法的执行过程，并理解它们对性能的影响。
Computational thinking skills — decomposition, pattern recognition and abstraction — underpin the design of efficient algorithms.
计算思维技能——分解、模式识别和抽象——是设计高效算法的基础。
11. Programming Fundamentals | 编程基础
Year 10 learners are expected to write, trace and debug programs using a high‑level language (often Python). Core constructs include sequence, selection (if‑elif‑else) and iteration (while and for loops). You must use meaningful variable names and comments to explain your logic.
Year 10 学生需要会使用高级语言（常为 Python）编写、跟踪与调试程序。核心结构包括顺序、选择（if‑elif‑else）和迭代（while 和 for 循环）。你必须使用有意义的变量名和注释来解释逻辑。
rograms also employ arithmetic operators (+, −, *, /, MOD, DIV), relational operators (==, !=, <, >, <=, >=) and logical operators (AND, OR, NOT).
需要掌握的数据类型：整数、浮点数、字符串、布尔值以及数组/列表。程序中还使用算术运算符（+、−、*、/、MOD、DIV）、关系运算符（==、!=、<、>、<=、>=）和逻辑运算符（AND、OR、NOT）。
Subprograms (functions and procedures) allow code reuse. A function returns a value, while a procedure performs a task without returning a value. Parameters and arguments let you pass data into a subprogram.
子程序（函数和过程）可实现代码重用。函数会返回一个值，而过程执行任务但不返回值。参数与实参用于将数据传入子程序。
Published by TutorHao | Computer Science Revision Series | aleveler.com
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