📚 IGCSE AQA Computer Science: High-Frequency Topics Summary | IGCSE AQA 计算机：高频考点总结
This revision guide consolidates the most frequently examined topics in the IGCSE AQA Computer Science specification. Each key concept is explained briefly in English paired with its Chinese translation, covering data representation, hardware, networks, programming, algorithms, logic, databases, and societal impacts to help you focus your final review.
这篇复习指南整理了IGCSE AQA 计算机科学考试中最常出现的高频考点。每个关键概念先用英文简要说明，再配以中文翻译，覆盖数据表示、硬件、网络、编程、算法、逻辑、数据库和社会影响等核心领域，帮助你在最后冲刺阶段高效复习。
1. Binary and Hexadecimal Number Systems | 二进制与十六进制数制
Computers process data using binary (base-2) because transistors have two stable states (on/off). A single binary digit is called a bit. Denary (base-10) numbers can be converted to binary by repeated division by 2, recording the remainders. To convert binary to denary, add the powers of 2 for each position containing a 1.
计算机使用二进制（基数为2）处理数据，因为晶体管只有开和关两种稳定状态。一个二进制位称为比特。十进制数可以通过反复除以2并记录余数的方式转换为二进制。将二进制每一位的1乘以对应的2的幂再求和，即可转回十进制。
Hexadecimal (base-16) uses digits 0–9 and letters A–F, providing a compact way to represent binary values. Each hex digit corresponds to a nibble (4 bits). Conversion is simple: group binary bits into nibbles and replace each with its hex equivalent.
十六进制（基数为16）使用数字0–9和字母A–F，能以更紧凑的形式表示二进制值。每个十六进制位对应4个二进制位（一个“半字节”）。转换方法很简单：将二进制位按4位一组划分，每组转换为对应的十六进制符号。
1101 1010₂ = DA₁₆
Binary addition follows simple rules: 0+0=0, 1+0=1, 1+1=0 carry 1, and 1+1+carry=1 carry 1. An overflow error occurs when the result exceeds the allocated number of bits and the extra carry bit cannot be stored.
二进制加法遵循简单规则：0+0=0，1+0=1，1+1=0进1，1+1+进位=1进1。当运算结果超出分配给数字的位数，多出的进位无法保存时，就会发生溢出错误。
1010₂ + 0110₂ = 1 0000₂ (overflow in a 4‑bit system)
2. Data Representation: Text, Images and Sound | 数据表示：文本、图像与声音
Text is stored using character codes. ASCII uses 7 bits (128 characters), while extended ASCII uses 8 bits. Unicode can represent characters from all major languages using up to 32 bits per character, ensuring global compatibility.
文本通过字符编码存储。ASCII使用7位（128个字符），扩展ASCII使用8位。Unicode能为所有主要语言的字符提供编码，每个字符最多可使用32位，保证全球兼容性。
Images are represented as a grid of pixels. Each pixel’s colour is stored as a binary number. The colour depth (bits per pixel) determines how many colours can be displayed. Resolution is the total number of pixels; higher resolution and colour depth give better quality but larger file sizes.
图像用像素网格表示。每个像素的颜色以二进制数存储。颜色深度（每像素位数）决定了可显示的颜色数量。分辨率即像素总数；分辨率和颜色深度越高，图像质量越好，但文件也越大。
Sound is analogue; to store it digitally, the sound wave is sampled at regular intervals. The sample rate (Hz) and bit depth affect quality and file size. Higher sample rate and bit depth capture more detail but increase storage needs.
声音是模拟信号，要数字化存储就需要按固定间隔对声波进行采样。采样率（Hz）和位深度影响音质和文件大小。采样率和位深度越高，捕捉的细节越多，但所需的存储空间也越大。
3. CPU and the Fetch-Decode-Execute Cycle | CPU 与取指-译码-执行周期
The Central Processing Unit (CPU) processes instructions. It contains the Control Unit (CU) to manage the flow of data, the Arithmetic Logic Unit (ALU) to perform calculations and logical operations, and registers (e.g. Program Counter, Memory Address Register, Memory Data Register, Accumulator).
中央处理器（CPU）负责执行指令。它包含控制单元（CU）管理数据流、算术逻辑单元（ALU）执行计算和逻辑操作，以及多个寄存器（如程序计数器、存储器地址寄存器、存储器数据寄存器和累加器）。
The fetch-decode-execute cycle is the basic operation of a CPU: fetch the next instruction from memory (address in PC) into the MDR through the MAR; decode the instruction; execute it, often involving the ALU and accumulator. The cycle repeats continuously, with the PC incrementing each time.
取指-译码-执行周期是CPU的基本工作流程：根据程序计数器中的地址从内存取指令，通过地址寄存器将指令加载到数据寄存器；译码；执行，通常涉及ALU和累加器。周期不断重复，每次程序计数器都会递增。
Factors affecting CPU performance include clock speed (cycles per second), number of cores (parallel processing), and cache size (fast memory that stores frequently used data).
影响CPU性能的因素包括时钟频率（每秒周期数）、核心数量（并行处理能力）和高速缓存大小（用于存储常用数据的快速内存）。
4. Primary and Secondary Storage | 主存与辅助存储
Primary storage consists of RAM and ROM. RAM is volatile, used for data and programs currently in use. ROM is non-volatile and stores the boot program (firmware). Virtual memory uses part of the hard disk as an extension of RAM when physical RAM is full.
主存储器包括RAM和ROM。RAM是易失性的，用于存放当前正在使用的数据和程序。ROM是非易失性的，存储引导程序（固件）。当物理RAM不足时，虚拟内存会使用硬盘的一部分作为RAM的扩展。
Secondary storage is non-volatile and retains data when power is off. Magnetic hard disk drives (HDD) are cost-effective but slower. Solid-state drives (SSD) use flash memory, are faster and more durable but cost more per GB. Optical media (CD, DVD) are now less common but can be used for distribution.
辅助存储器是非易失性的，断电后仍能保留数据。磁性硬盘（HDD）成本低，但速度较慢。固态硬盘（SSD）使用闪存，速度快且耐用，但每GB成本更高。光盘（CD、DVD）现已不太常用，但仍可用于分发数据。
5. Database Concepts and SQL | 数据库概念与 SQL
A relational database stores data in tables, where each table consists of records (rows) and fields (columns). A primary key uniquely identifies each record. A foreign key links tables together.
关系数据库将数据存储在表中，每张表由记录（行）和字段（列）组成。主键惟一标识每条记录，外键用于将表关联起来。
SQL (Structured Query Language) is used to manage and query data. A common query: SELECT field1, field2 FROM table WHERE condition; Results can be sorted with ORDER BY.
SQL（结构化查询语言）用于管理和查询数据。常见查询：SELECT field1, field2 FROM table WHERE condition; 可以用 ORDER BY 对结果排序。
| StudentID | Name | Year |
|---|---|---|
| 001 | Alice | 10 |
| 002 | Bob | 11 |
SELECT Name FROM Students WHERE Year = 10;
This retrieves the names of all students in Year 10. Mastering simple SELECT statements with AND/OR conditions is essential for the exam.
这条语句会检索所有10年级学生的姓名。掌握包含AND/OR条件的简单SELECT语句对考试至关重要。
6. Networks and Topologies | 网络与拓扑
A LAN (Local Area Network) covers a small geographical area, often using Ethernet or Wi-Fi. A WAN (Wide Area Network) connects networks over large distances, such as the Internet.
局域网（LAN）覆盖较小的地理区域，常使用以太网或Wi-Fi。广域网（WAN）在较大范围内连接多个网络，如互联网。
In a client-server model, a central server provides services or resources to clients. In a peer-to-peer network, all devices share resources equally without a dedicated server.
在客户-服务器模型中，中央服务器为客户机提供服务或资源。在对等网络中，所有设备平等共享资源，无需专用服务器。
A star topology connects all nodes to a central switch; it is reliable because one cable failure does not affect others. A mesh
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