Core Knowledge Review for CAIE Year 10 Computer Science | Year 10 CAIE 计算机：核心知识点梳理

📚 Core Knowledge Review for CAIE Year 10 Computer Science | Year 10 CAIE 计算机：核心知识点梳理

As Year 10 students progress through the CAIE IGCSE Computer Science course, building a solid foundation of core concepts is essential. This article reviews key topics including data representation, logic, hardware, software, algorithms, programming, databases, and security, providing clear explanations and practical examples to reinforce understanding.

随着Year 10学生深入学习CAIE IGCSE计算机科学课程，打下扎实的核心概念基础至关重要。本文梳理了数据表示、逻辑、硬件、软件、算法、编程、数据库和安全等关键主题，提供清晰的解释和实用示例，以巩固理解。

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1. Number Systems & Data Representation | 数制与数据表示

Computers use binary (base-2) because digital circuits have two states: on (1) and off (0). A single binary digit is a bit; 8 bits form a byte. To shorten long binary strings, we use hexadecimal (base-16), which groups bits into nibbles of four.

计算机使用二进制（基数为2），因为数字电路只有开(1)和关(0)两种状态。一个二进制位称为比特，8个比特构成1个字节。为了缩短二进制的长度，我们使用十六进制（基数为16），每4位一组进行转换。

¹ 2⁰ . For 1101₂, the value is 1×8 + 1×4 + 0×2 + 1×1 = 13. In hex, digits A-F represent 10– o binary: replace each hex digit with the equivalent 4-bit nibble, e.g. A3₁₆ = 1010 0011₂.

二进制中每一位的权值是2的幂：... 2³ 2² 2¹ 2⁰ 。以1101₂为例，其值为1×8 + 1×4 + 0×2 + 1×1 = 13。十六进制中字母A-F分别代表10–15。十六进制转二进制只需将每位十六进制替换为等值的4位二进制，如 A3₁₆ = 1010 0011₂。

the result exceeds the available bit width. Logical shifts move bits left or right; a left shift multiplies by 2, and a right shift divides by 2 (integer division).

二进制加法规则简单：0+0=0，0+1=1，1+1=0进1，1+1+进位=1进1。当结果超过可用位宽时会产生溢出。逻辑移位将位向左或向右移动；左移一位相当于乘以2，右移一位相当于除以2（整数除法）。

Negative integers are often stored using two's complement. To obtain the two's complement of a number: invert all bits (one' 1111010₂, add 1 → 11111011₂. The most significant bit acts as the sign bit (1 = negative).

负整数通常以补码(two's complement)形式存储。求一个数的补码：先将所有位取反（反码），然后加1。例如，8位中表示-5：从00000101₂开始，取反得11111010₂，加1得11111011₂。最高位充当符号位（1表示负数）。

Text is stored by assigning a numeric code to each character. ASCII uses 7 bits (extended ASCII 8 bits), covering 128/256 characters. Unicode provides a much larger range (e.g. UTF-8, UTF-16), supporting international scripts and emojis.

文本通过为每个字符分配数值码来存储。ASCII使用7位（扩展ASCII为8位），可表示128/256个字符。Unicode则提供更广的字符集（如UTF-8、UTF-16），支持各国文字和表情符号。

s: 1 bit = 2 colours, 8 bits = 256 colours, 24 bits ≈ 16.7 million colours. Image file size (bits) = resolution width × height × colour depth.

图像由像素构成。分辨率是像素数量（如1920×1080），色彩深度表示每像素的位数。位越多颜色越多：1位=2种颜色，8位=256种颜色，24位≈1670万种颜色。图像文件大小（比特）= 宽×高×色彩深度。

re higher frequencies. Bit depth determines the number of possible amplitude levels (2ⁿ levels). Stereo sound uses two channels. File size (bits) = sample rate × bit depth × duration × channels.

声音通过定期对模拟信号采样来录制。采样率（Hz）是每秒采样次数，越高能捕获的频率越高。采样精度决定振幅等级的数目（2ⁿ个等级）。立体声有两个声道。文件大小（比特）= 采样率 × 精度 × 时长 × 声道数。

Compression reduces file size. Lossy compression (e.g. JPEG, MP3) permanently removes some data to achieve smaller sizes; lossless compression (e.g. PNG, ZIP) preserves all original data and reconstructs the exact file. Run-length encoding is a simple lossless method for repeating values.

压缩可以减小文件大小。有损压缩（如JPEG、MP3）会永久移除部分数据以换取更小的体积；无损压缩（如PNG、ZIP）保留全部原始数据并能完全重建。游程编码是一种简单的无损方法，适用于重复值。

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2. Logic Gates & Boolean Algebra | 逻辑门与布尔代数

A logic gate is an electronic circuit that performs a Boolean function. The most basic gates are NOT, AND, OR, NAND, NOR, and XOR. Each gate can be described by its symbol, Boolean expression, and truth table.

逻辑门是实现布尔函数的电子电路。最基本的门有NOT、AND、OR、NAND、NOR和XOR。每个门都可用符号、布尔表达式和真值表来描述。

Gate	Symbol (Boolean)	Truth Table
NOT	Q = A̅ or Q = NOT A	A=0 → Q=1; A=1 → Q=0
AND	Q = A · B	All inputs 1 → Q=1; else Q=0
OR	Q = A + B	At least one input 1 → Q=1
NAND	Q = A ⊼ B (NOT AND)	Opposite of AND
NOR	Q = A ⊽ B (NOT OR)	Opposite of OR
XOR	Q = A ⊕ B	Inputs different → Q=1; same → Q=0

uit helps verify its behaviour. Boolean algebra can simplify circuits: for instance, A + A̅·B = A + B.

逻辑电路由多个门组合而成。从输入端开始追踪信号，直到最终输出。为电路建立真值表有助于验证其行为。布尔代数可以化简电路，例如 A + A̅·B = A + B。

xample, a system that outputs 1 when two sensors A and B are both high OR when a master switch M is high: Q = (A AND B) OR M.

常见的考题是从文字描述设计逻辑系统。写出布尔表达式，尽可能简化，然后画出门电路图。例如，系统在传感器A与B同时为高点或主开关M为高点时输出1：Q = (A AND B) OR M。

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3. Computer Architecture | 计算机体系结构

The Von Neumann architecture is the foundation of most modern computers. It stores both data and instructions in the same memory, using a single bus system. Key components are the Central Processing Unit (CPU), memory, input/output controllers, and system buses (address, data, control).

冯·诺依曼架构是现代大多数计算机的基础。它将数据和指令存放在同一个内存中，使用单一总线系统。核心组件包括中央处理器(CPU)、内存、输入/输出控制器和系统总线（地址总线、数据总线、控制总线）。

The CPU consists of the Arithmetic Logic Unit (ALU), which performs calculations and logical operations, and the Control Unit (CU), which directs the fetch‑decode‑execute cycle. Important registers include the Program Counter (PC), Memory Address Register (MAR), Memory Data Register (MDR), Current Instruction Register (CIR), and Accumulator (ACC).

CPU由执行运算和逻辑操作的算术逻辑单元(ALU)和指挥取指-译码-执行周期的控制单元(CU)组成。重要的寄存器包括程序计数器(PC)、内存地址寄存器(MAR)、内存数据寄存器(MDR)、当前指令寄存器(CIR)和累加器(ACC)。

The fetch‑decode‑execute cycle works as follows: PC holds the address of the next instruction; this address is copied to MAR and the instruction is fetched into MDR, then moved to CIR. The CU decodes the instruction and may involve the ALU, then the result is stored, and PC is incremented. This cycle repeats billions of times per second (clock speed).

取指-译码-执行周期的工作过程：PC保存下一条指令的地址；该地址被复制到MAR，指令被取入MDR，再移入CIR。CU对指令译码，可能需要ALU参与，然后结果被保存，PC递增。该周期每秒重复数十亿次（时钟速度）。

There are three system buses: the address bus (carries memory locations, unidirectional), the data bus (carries data between CPU and memory/IO, bidirectional), and the control bus (carries control signals like read/write). Bus width affects performance—more lines mean more data transferred per cycle.

有三条系统总线：地址总线（传送内存地址，单向）、数据总线（在CPU与内存/IO之间传输数据，双向）和控制总线（传输读写等控制信号）。总线宽度影响性能——线数越多，每个周期传输的数据量越大。

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4. Memory & Storage | 内存与存储

Primary memory is directly accessed by the CPU. RAM (Random Access Memory) is volatile, holds data and instructions currently in use, and can be read from and written to. ROM (Read Only Memory) is non-volatile, stores firmware/BIOS, and is read-only during normal operation.

主存由CPU直接访问。RAM（随机存取存储器）是易失性的，存放当前使用的数据和指令，可读写。ROM（只读存储器）是非易失性的，存放固件/BIOS，正常运行时只能读取。

Virtual memory uses part of the hard disk/SSD when RAM is full. It allows larger programs to run but is much slower because disk access is slower than RAM. The operating system swaps pages between RAM and secondary storage.

当RAM不足时，虚拟内存会使用硬盘/SSD的一部分空间。它允许运行大型程序，但由于磁盘访问比RAM慢很多，速度显著下降。操作系统会在RAM和辅存之间交换页面。

Secondary storage provides permanent, non-volatile storage. Magnetic hard disk drives (HDD) use spinning platters and read/write heads. Solid-state drives (SSD) use flash memory with no moving parts, making them faster, quieter, and more durable. Optical discs (CD, DVD, Blu-ray) use laser technology. Cloud storage keeps data on remote servers accessed via the internet.

辅助存储器提供永久性的非易失存储。机械硬盘(HDD)使用旋转的盘片和读写磁头。固态硬盘(SSD)采用闪存，无机械部件，因此更快、更安静且更耐用。光盘（CD、DVD、蓝光）使用激光技术。云存储将数据保存在远程服务器上，可通过网络访问。

e (TB ≈ 10¹²). Note the difference between decimal (10³) and binary (2¹⁰ = 1024) prefixes.

存储容量以字节为单位：千字节(KB ≈ 10³字节)、兆字节(MB ≈ 10⁶)、吉字节(GB ≈ 10⁹)、太字节(TB ≈ 10¹²)。注意十进制(10³)和二进制(2¹⁰=1024)前缀的区别。

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5. Software & Operating Systems | 软件与操作系统

Software is divided into system software and application software. System software includes the operating system (OS), utilities, and device drivers. Application software performs specific tasks for users, such as word processors, web browsers, and games.

软件分为系统软件和应用软件。系统软件包括操作系统(OS)、实用工具和设备驱动程序。应用软件为用户执行特定任务，如文字处理软件、网页浏览器和游戏。

The OS acts as an interface between the user and hardware. Its key functions: memory management (allocates RAM to processes), multitasking (switches rapidly between tasks), file management (organises files in folders), peripheral management (driver interaction), security (user accounts, access control), and providing a user interface (GUI or CLI).

OS是用户与硬件之间的接口。其主要功能包括：内存管理（为进程分配RAM）、多任务处理（在各任务间快速切换）、文件管理（以文件夹组织文件）、外设管理（与驱动程序交互）、安全（用户账户、访问控制）以及提供用户界面（GUI或CLI）。

Interrupts are signals sent by hardware or software that cause the OS to temporarily halt the current process, save its state, and run an interrupt service routine. This allows the system to respond to events such as a mouse click, keystroke, or completion of I/O.

中断是由硬件或软件发出的信号，它使操作系统暂停当前进程、保存状态并执行中断服务程序。这使得系统能响应鼠标点击、按键或I/O完成等事件。

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6. Algorithms & Problem-solving | 算法与问题解决

An algorithm is a step-by-step sequence of instructions to solve a problem. It is designed before coding, often using pseudocode or flowcharts. Pseudocode uses structured English-like statements; flowcharts use standard symbols: oval for start/end, parallelogram for input/output, rectangle for process, diamond for decision, arrows for flow.

算法是逐步执行的指令序列，用以解决问题。它在编码前设计，常使用伪代码或流程图。伪代码采用结构化的类英语语句；流程图使用标准符号：椭圆表示开始/结束，平行四边形表示输入/输出，矩形表示处理过程，菱形表示判断，箭头表示流向。

Key constructs: sequence (steps in order), selection (IF...THEN...ELSE...ENDIF, CASE OF), and iteration (FOR loop, WHILE loop, REPEAT...UNTIL). A trace table tracks how variable values change as an algorithm runs, helping to identify logic errors.

基本结构：顺序（按步骤执行）、选择（IF...THEN...ELSE...ENDIF、CASE OF）和迭代（FOR循环、WHILE循环、REPEAT...UNTIL）。跟踪表可以记录算法执行时变量值的变化，有助于发现逻辑错误。

Validation checks whether data is sensible and reasonable (e.g. range check, type check, presence check). Verification ensures data has been entered correctly (e.g. double entry, visual check). Test data should include normal, boundary (extreme), and erroneous values.

验证检查数据是否合理（如范围检查、类型检查、存在性检查）。校验确保数据被正确输入（如双重录入、人工核对）。测试数据应包括正常值、边界值（极端值）和错误值。

Standard searching algorithms: linear search examines each element in turn until the target is found (works on unordered lists). Binary search works on ordered lists by repeatedly dividing the search space in half; it is more efficient but requires sorted data.

标准搜索算法：线性搜索逐个检查元素直到找到目标（适用于无序列表）。二分搜索适用于有序列表，通过反复将搜索区间折半进行，效率更高但需要数据已排序。

Standard sorting: bubble sort repeatedly swaps adjacent elements if they are in the wrong order, moving the largest element to the end

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