📚 Year 10 Cambridge IGCSE Computer Science Summer Bridging Course | 剑桥 IGCSE 计算机科学 Year 10 暑期衔接课程
Welcome to the Year 10 Cambridge IGCSE Computer Science summer bridging programme. This course is designed to prepare you for the two-year IGCSE 0478 syllabus by introducing essential concepts from both the theoretical and the practical aspects of computing. Over the coming weeks you will meet number systems, hardware, networks, security, algorithms and logic gates - the same topics that will carry you through two years of lessons and two final written papers.
欢迎参加 Year 10 剑桥 IGCSE 计算机科学暑期衔接课程。本课程旨在通过介绍计算机科学的理论与实践基础，帮助你提前适应 IGCSE 0478 两年制课程的学习。在接下来的几周里，你将接触数制、硬件、网络、安全、算法与逻辑门——这些正是贯穿两年课堂以及两场最终笔试的核心主题。
Read this guide as a roadmap rather than a textbook. Each section states what the topic is, why the examiner cares about it, and what you should be able to do by the end of the summer. Work through the sections in order, attempt the worked examples on paper, and keep a single notebook for all conversions, truth tables and pseudocode - you will reuse it in September.
请把本文当作一张路线图，而不是一本教科书。每一节都会说明该主题是什么、考官为什么重视它，以及你在暑假结束时应该掌握到什么程度。建议按顺序阅读，把例题亲手算一遍，并用同一个笔记本记录所有进制转换、真值表和伪代码——九月开学后你还会继续用到它。
1. Course Overview and Importance | 课程概览与重要性
The Cambridge IGCSE Computer Science qualification develops computational thinking, problem solving and programming skills. The syllabus covers topics such as data representation, computer systems, networks, security and algorithms, and it asks you to think like a machine as well as like a problem solver.
剑桥 IGCSE 计算机科学课程培养计算思维、问题解决和编程能力。课程大纲涵盖数据表示、计算机系统、网络、安全以及算法等主题，它要求你既能像机器一样思考，也能像问题解决者一样思考。
Starting early with a bridging course helps you build confidence in binary arithmetic, logic gates and basic programming before the term begins. Familiarity with these foundations will allow you to focus on higher-order skills during the school year.
通过衔接课程提前学习，你可以在开学前就掌握二进制运算、逻辑门和基础编程等核心知识，建立信心。熟悉这些基础能让你在学年中更专注于高阶思维能力的培养。
One warning before you begin: IGCSE Computer Science is not the same subject as ICT. ICT teaches you how to use existing software; Computer Science teaches you how that software and the hardware beneath it actually work, and how to design your own solutions. Expect to convert numbers by hand, draw logic circuits and write code that runs.
开始之前有一点需要提醒：IGCSE 计算机科学并不等同于 ICT。ICT 教你如何使用现成的软件，而计算机科学教你这些软件及其底层硬件究竟如何运作，以及如何设计自己的解决方案。你要准备好手算进制转换、绘制逻辑电路，并写出能够真正运行的代码。
2. How the IGCSE 0478 Syllabus and Exams Are Structured | IGCSE 0478 大纲与考试结构
The 0478 specification is split into two written papers of equal weight. There is no coursework and no practical examination, so every mark you earn comes from a pen-and-paper answer. Knowing this on day one changes how you revise: you must be able to explain a concept clearly, not merely use it.
0478 考纲分为两场权重相同的笔试，没有课程作业，也没有实操考试，因此你拿到的每一分都来自纸笔作答。开学第一天就明白这一点会改变你的复习方式：你必须能够清晰地解释一个概念，而不只是会用它。
| Paper | 试卷 | Content | 内容 | Duration | 时长 | Marks | 分值 | Weighting | 权重 |
|---|---|---|---|---|
| Paper 1 | Computer Systems | 计算机系统 | 1 h 45 min | 75 | 50% |
| Paper 2 | Algorithms, Programming and Logic | 算法、编程与逻辑 | 1 h 45 min | 75 | 50% |
The theoretical content is organised into six areas: data representation, data transmission, hardware, software, the internet and its uses, and automated and emerging technologies. The practical content adds algorithm design and problem solving, programming, databases and Boolean logic.
理论部分分为六大板块：数据表示、数据传输、硬件、软件、互联网及其应用，以及自动化与新兴技术。实践部分则增加算法设计与问题求解、程序设计、数据库以及布尔逻辑。
Because both papers are written, command words matter enormously. "State" needs a short factual answer; "describe" needs a point plus a detail; "explain" needs a cause-and-effect link, usually signalled by the word "because". Train yourself to count the marks and give exactly that many distinct points.
由于两场考试都是笔试，题干中的指令词极其重要。"State（写出）"只需要简短的事实答案；"describe（描述）"需要要点加细节；"explain（解释）"需要因果关联，通常要用"because"来体现。请训练自己先数字数中的分值，然后给出恰好相同数量的独立要点。
3. Data Representation: Binary and Hexadecimal | 数据表示：二进制与十六进制
Binary is a base-2 number system that uses only the digits 0 and 1. All digital data inside a computer is stored and processed in binary form because electronic circuits have two stable states (on and off), which map naturally onto the symbols 1 and 0.
二进制是一种只使用数字 0 和 1 的基数为 2 的计数系统。计算机内部所有数字数据都以二进制形式存储和处理，因为电子电路有两种稳定状态（开与关），它们可以自然地对应到符号 1 和 0。
Denary is base 10, binary is base 2, and hexadecimal is base 16, using the digits 0 to 9 followed by A to F for the values 10 to 15. The key relationship to memorise is that one hexadecimal digit represents exactly four bits (one nibble), so two hexadecimal digits represent one byte.
十进制是基数 10，二进制是基数 2，十六进制是基数 16，使用数字 0 到 9 以及 A 到 F 表示数值 10 到 15。必须牢记的关键关系是：一个十六进制位恰好表示四位二进制（一个 nibble），因此两个十六进制位表示一个字节。
That is exactly why memory addresses, MAC addresses and colour codes such as #FF8800 are written in hexadecimal. A 32-bit address becomes eight readable characters instead of thirty-two, which is far shorter for humans and much less likely to be misread or mis-typed.
这也正是内存地址、MAC 地址以及 #FF8800 这类颜色代码使用十六进制书写的原因。一个 32 位地址会变成八个可读字符，而不是三十二个字符，对人类而言更简短，也更不容易看错或输错。
Denary 202 = 11001010₂ = CA₁₆
Work it out in both directions: split 202 into place values 128 + 64 + 8 + 2 to get 11001010, then split the binary into two nibbles, 1100 and 1010, giving C and A. Reversing the process is the same skill used backwards, and examiners test both directions every year.
请双向练习：把 202 拆成位值 128 + 64 + 8 + 2，得到 11001010；再把二进制分成两个 nibble，即 1100 和 1010，分别对应 C 和 A。反向过程用的是同一套技能，只是倒过来做，而考官每年都会考查这两个方向。
4. Binary Arithmetic, Shifts and Overflow | 二进制运算、移位与溢出
+ 0 = 0, 0 + 1 = 1, 1 + 1 = 10 (write 0, carry 1), and 1 + 1 + 1 = 11 (write 1, carry 1).
二进制加法要从右往左逐列进行，当某一列之和达到 2 时就进位。规则很短：0 + 0 = 0，0 + 1 = 1，1 + 1 = 10（写 0，进 1），1 + 1 + 1 = 11（写 1，进 1）。
1011 0110 (182) + 0101 1101 (93) = 1 0001 0011 (275) → overflow
g it is a standard exam question. An 8-bit register can hold values from 0 to 255 only, since 2⁸ = 256 distinct patterns including zero.
真实的和是 275，需要九位二进制，但只有八位可用。那个无处安放的额外位就叫溢出（overflow），识别它是标准的考题。一个 8 位寄存器只能保存 0 到 255 之间的值，因为 2⁸ = 256 个不同组合中包含了零。
A logical shift moves every bit one or more places in the same direction, filling the empty positions with zeros. A logical shift left by one place doubles the value; a logical shift right by one place halves it, discarding any bit that falls off the end.
逻辑移位会让每一位朝同一方向移动一位或多位，空出的位置用 0 填充。左移一位相当于把数值乘以 2，右移一位相当于除以 2，移出末端的位会被丢弃。
0000 1010 (10) → left shift → 0001 0100 (20) → right shift → 0000 1010 (10)
Two's complement, where the leftmost bit acts as a sign bit, is introduced at A Level rather than at IGCSE, but knowing that the most significant bit can carry a negative weight gives you a useful head start and explains why overflow behaves differently for signed values.
用最高位作为符号位的"二进制补码"是 A Level 的内容，IGCSE 阶段并不要求，但了解最高有效位可以带有负权重，会给你一个有用的先手优势，也能解释为什么有符号数的溢出行为与无符号数不同。
5. Representing Text, Sound and Images | 文本、声音与图像的表示
Characters are stored using character codes. ASCII uses 7 bits and defines 128 characters, which is enough for English letters, digits and punctuation. Extended ASCII uses 8 bits for 256 characters, and Unicode uses 16 bits or more, which is enough to represent every writing system in the world.
字符通过字符编码来存储。ASCII 使用 7 位，定义了 128 个字符，足以容纳英文字母、数字和标点。扩展 ASCII 使用 8 位，可表示 256 个字符；而 Unicode 使用 16 位甚至更多，足以表示世界上所有书写系统。
Sound is analogue, so it must be sampled to become digital. The sample rate is how many samples are taken per second, and the sample resolution (bit depth) is how many bits are used for each sample. Higher values for either give better quality and larger files.
声音是模拟信号，必须经过采样才能数字化。采样率指每秒采样的次数，采样分辨率（位深度）指每个采样点使用的位数。两者取值越高，音质越好，文件也越大。
und file size (bits) = sample rate × sample resolution × duration (s) × number of channels
A bitmap image is a grid of pixels. Its resolution is the number of pixels across and down, and its colour depth is the number of bits used per pixel. Doubling the colour depth doubles the file size without changing the printed dimensions.
位图图像是由像素组成的网格。它的分辨率指横向和纵向的像素数量，色深指每个像素使用的位数。色深翻倍会让文件大小翻倍，而图像的打印尺寸不变。
Image file size (bits) = width (px) × height (px) × colour depth (bits)
0 bits, which is 1 440 000 bytes, or about 1.37 MiB. Always divide by 8 to convert bits to bytes, and show your working - method marks are awarded even when the final arithmetic slips.
例如，一幅 800
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