IGCSE Computer Science: Detailed Explanation of Typical Exam Questions | IGCSE 计算机：典型例题详解

📚 IGCSE Computer Science: Detailed Explanation of Typical Exam Questions | IGCSE 计算机：典型例题详解

This article walks through typical IGCSE Computer Science exam questions, providing step-by-step solutions and explanations. Each section focuses on a specific topic area with worked examples to help you master the key concepts and examination techniques.

本文精选 IGCSE 计算机科学典型考题，逐步详解解题思路与方法。每个小节聚焦一个主题，配合典型例题和详细解析，帮助你扎实掌握核心考点与答题技巧。

1. Number Systems: Binary and Hexadecimal Conversions | 数制：二进制与十六进制转换

Example: Convert the denary number 156 into an 8‑bit binary number and then into hexadecimal. Show all working.

例题：将十进制数 156 转换为 8 位二进制数，再转换为十六进制。写出完整步骤。

Solution: To convert 156 to binary, repeatedly divide by 2 and record the remainders: 156 ÷ 2 = 78 rem 0, 78 ÷ 2 = 39 rem 0, 39 ÷ 2 = 19 rem 1, 19 ÷ 2 = 9 rem 1, 9 ÷ 2 = 4 rem 1, 4 ÷ 2 = 2 rem 0, 2 ÷ 2 = 1 rem 0, 1 ÷ 2 = 0 rem 1. Reading remainders in reverse gives 10011100. As an 8‑bit binary, we pad to eight digits: 10011100. To convert to hex, split binary into nibbles: 1001 1100. 1001 is 9 in hex, 1100 is C in hex. Therefore, 156 in hexadecimal is 9C.

解答：将 156 转换为二进制，连续除以 2 并记录余数：156 ÷ 2 = 78 余 0，78 ÷ 2 = 39 余 0，39 ÷ 2 = 19 余 1，19 ÷ 2 = 9 余 1，9 ÷ 2 = 4 余 1，4 ÷ 2 = 2 余 0，2 ÷ 2 = 1 余 0，1 ÷ 2 = 0 余 1。从下往上读出余数得到 10011100。作为 8 位二进制，已经是 8 位。转换为十六进制时，将二进制拆分为半字节：1001 和 1100。1001 对应十六进制 9，1100 对应 C。因此 156 的十六进制为 9C。

2. Binary Addition and Overflow | 二进制加法与溢出

Example: Perform the binary addition of 10110110 + 01101001 using 8‑bit registers. State whether an overflow occurs and explain why.

例题：在 8 位寄存器中进行二进制加法 10110110 + 01101001。判断是否发生溢出，并解释原因。

Solution: Add column by column from right to left, carrying where necessary. 0+1=1, 1+0=1, 1+0=1, 0+1=1, 1+0=1, 1+1=0 carry 1, 1+1+carried 1 = 1 carry 1, 1+0+carried 1 = 0 carry 1. The result is 00011111 with a carry-out of 1. In 8‑bit unsigned representation, a carry-out from the most significant bit indicates an overflow because the sum (438) exceeds the maximum value 255. For signed two’s complement interpretation, adding a negative and positive number should not overflow, but here the carry into and out of the sign bit are both 1, so no signed overflow occurs. The answer depends on the context: for unsigned arithmetic, overflow happens; for signed two’s complement, it does not. The typical IGCSE examination expects you to identify an overflow when the result is outside the representable range of the given number of bits.

解答：从右向左逐列相加，需要时进位。0+1=1，1+0=1，1+0=1，0+1=1，1+0=1，1+1=0 进 1，1+1+进位 1 = 1 进 1，1+0+进位 1 = 0 进 1。结果为 00011111，最高位产生进位 1。在 8 位无符号表示中，最高位的进位代表溢出，因为和 (438) 超过了最大值 255。若解释为有符号补码，进位数入符号位与出符号位均为 1，因此没有有符号溢出。考试中通常要求根据指定位数是否超出范围来判断溢出，无符号加法下产生溢出。

3. Logic Gates and Truth Tables | 逻辑门与真值表

Example: A logic circuit has inputs A, B and output X given by X = (A AND B) OR (NOT C). Draw the truth table for all possible input combinations.

例题：某逻辑电路有输入 A、B、C，输出 X = (A AND B) OR (NOT C)。画出所有输入组合的真值表。

Solution: First list all eight combinations of A, B, C (0 or 1). Evaluate A AND B, NOT C, then the final OR. The truth table is shown below.

解答：首先列出 A、B、C 所有 8 种组合（0 或 1）。计算 A AND B、NOT C，再计算最终的 OR。真值表如下。

A	B	C	A AND B	NOT C	X
0	0	0	0	1	1
0	0	1	0	0	0
0	1	0	0	1	1
0	1	1	0	0	0
1	0	0	0	1	1
1	0	1	0	0	0
1	1	0	1	1	1
1	1	1	1	0	1

4. Simplifying Boolean Expressions | 布尔表达式化简

Example: Simplify the Boolean expression (A + B) • (A + ¬B) using Boolean algebra laws. State the laws used at each step.

例题：利用布尔代数定律化简表达式 (A + B) • (A + ¬B)，并注明每一步所使用的定律。

Solution: Apply the distributive law in reverse (or expand and then simplify). Expanding gives A•A + A•¬B + B•A + B•¬B. Since A•A = A, B•¬B = 0, we obtain A + A•¬B + A•B + 0 = A + A•¬B + A•B. Factor A out of the last two terms: A + A•(¬B + B). Since ¬B + B = 1, this becomes A + A•1 = A + A = A. Alternatively, by the absorption law: (A + B)(A + ¬B) = A + (B•¬B) = A + 0 = A. So the simplified expression is A.

解答：运用分配律展开：A•A + A•¬B + B•A + B•¬B。因为 A•A = A，B•¬B = 0，得到 A + A•¬B + A•B + 0 = A + A•¬B + A•B。提取公因子 A：A + A•(¬B + B)。¬B + B = 1，化简为 A + A•1 = A + A = A。另一种方法，利用吸收律直接得到 (A + B)(A + ¬B) = A + B•¬B = A + 0 = A。化简结果为 A。

5. Memory and Storage: RAM and ROM | 内存与存储：RAM 与 ROM

Example: Compare RAM and ROM in terms of volatility, speed, typical uses, and whether data can be written to them. Explain why a computer needs both types of memory.

例题：比较 RAM 和 ROM 在易失性、速度、典型用途以及是否可写入数据方面的区别。解释为什么计算机需要这两种内存。

Solution: RAM (Random Access Memory) is volatile, meaning it loses its contents when power is turned off. It is used to store data and programs currently in use, allowing fast read and write operations by the CPU. ROM (Read Only Memory) is non‑volatile, retaining data without power. It typically stores firmware, such as the BIOS, and its contents are written during manufacture and cannot be easily modified. Computers need RAM to provide working space for the processor and running applications, and ROM to hold the essential startup instructions that boot the system.

解答：RAM（随机存取存储器）是易失性的，断电后数据丢失。它用于存放当前正在使用的数据和程序，CPU 可高速读写。ROM（只读存储器）是非易失性的，断电后仍能保存数据，通常用于存储固件，如 BIOS，其内容在制造时写入且一般无法修改。计算机需要 RAM 提供处理器和运行程序所需的临时工作空间，同时需要 ROM 保存启动时必须的指令，保证系统能够引导。

6. Components of a Computer System: The CPU and Von Neumann Architecture | 计算机系统组件：CPU 与冯·诺依曼架构

Example: Describe the fetch‑decode‑execute cycle of a CPU. In your answer, refer to the program counter (PC), memory address register (MAR), memory data register (MDR), current instruction register (CIR), and the control unit (CU).

例题：描述 CPU 的取指—解码—执行周期。回答中需提及程序计数器 (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) 以及控制单元 (CU)。

Solution: The cycle begins with the PC holding the address of the next instruction. This address is copied to the MAR, and the PC increments. The control unit sends a read signal to memory; the instruction at that address is transferred to the MDR. The instruction is then moved to the CIR. The control unit decodes the instruction, determining what operation is required. It then executes the instruction by coordinating the appropriate components (e.g. ALU for arithmetic, or loading data from memory). Once execution is complete, the cycle repeats by fetching the next instruction from the address in the PC.

解答：周期开始时，PC 保存着下一条指令的地址。该地址被复制到 MAR，然后 PC 自增。CU 发出读取信号，内存中对应地址的指令被送入 MDR，接着指令被移入 CIR。CU 对指令进行解码，确定需要执行的操作。然后 CU 协调相应部件执行指令（例如将数据送入 ALU 进行运算，或从内存加载数据）。执行完毕后，再次从 PC 指向的地址取出下一条指令，重复此周期。

7. Network Topologies and Transmission Media | 网络拓扑与传输介质

Example: Compare a star network topology with a bus topology. Include one advantage and one disadvantage for each, and state a suitable transmission medium for a school network.

例题：比较星型拓扑与总线拓扑。分别列举一项优点和一项缺点，并说明适合学校网络的传输介质。

Solution: In a star topology, all devices connect to a central switch or hub. Advantage: if one cable fails, only that device is affected. Disadvantage: if the central switch fails, the whole network goes down. In a bus topology, all devices share a single backbone cable. Advantage: it requires less cable than star. Disadvantage: a break in the backbone cable disables the entire network. For a school network, twisted‑pair copper cable (e.g. Cat 6) is commonly used because it is cost‑effective and supports high‑speed Ethernet connections. Alternatively, fibre optic cable may be used for backbone connections requiring higher bandwidth.

解答：星型拓扑中，所有设备连接到中央交换机或集线器。优点：一条电缆损坏只影响对应设备。缺点：中央交换机故障会导致全网瘫痪。总线拓扑中，所有设备共享一条主干电缆。优点：比星型所需电缆少。缺点：主干电缆断开会致使整个网络无法通信。学校网络通常采用双绞线（如 Cat 6），性价比高且支持高速以太网。对于需要高带宽的主干连接，也可使用光纤。

8. Cyber Security Threats and Prevention Measures | 网络安全威胁与防范措施

Example: Explain the difference between malware and phishing attacks. For each, describe one method that users can adopt to reduce the risk.

例题：解释恶意软件与网络钓鱼攻击的区别。针对每种威胁，各描述一种用户可采取的防范措施。

Solution: Malware is malicious software (e.g. virus, worm, trojan) that infects a system to damage data or gain unauthorised access. Phishing is a social engineering attack where attackers trick users into revealing personal information (e.g. passwords, bank details) by posing as a trustworthy entity, often through fake emails or websites. To reduce malware risk, users should install and regularly update anti‑malware software. To avoid phishing, users should verify the sender’s email address and avoid clicking on suspicious links, or enable multi‑factor authentication.

解答：恶意软件（如病毒、蠕虫、木马）是旨在破坏数据或获取未授权访问的恶意代码。网络钓鱼是一种社会工程学攻击，攻击者伪装成可信实体，常通过伪造邮件或网站诱骗用户泄露个人信息（如密码、银行信息）。针对恶意软件，用户应安装并及时更新反恶意软件。防范网络钓鱼，用户应核实发件人邮箱地址、不点击可疑链接，或启用多因素认证。

9. Algorithm Design and Pseudocode: Loop and Accumulation | 算法设计与伪代码：循环求和

Example: Study the following pseudocode and state the final value of total.

total ← 0
FOR i ← 1 TO 10
    total ← total + i
ENDFOR
OUTPUT total

例题：阅读以下伪代码，写出变量 total 的最终值。

Solution: The loop iterates i from 1 to 10 inclusive. Initially total = 0. In each iteration, total increases by i. This computes the sum of integers 1 to 10: 1+2+3+…+10 = 55. The final output is 55.

解答：循环变量 i 从 1 到 10（包含 10）。初始 total = 0，每次迭代 total 增加 i。该算法计算 1 到 10 的整数之和：1+2+3+…+10 = 55。最终输出为 55。

10. Database Querying with SQL: SELECT Statement | 数据库 SQL 查询：SELECT 语句

Example: A table named Students contains the fields StudentID, Name, Class, Age. Write an SQL statement to retrieve the Name and Class of all students whose Age is greater than 15.

例题：已知表 Students 包含字段 StudentID、Name、Class 和 Age。写出 SQL 语句，检索所有年龄大于 15 的学生的姓名和班级。

Solution: The required SQL uses SELECT to specify the columns Name and Class, FROM Students, with a WHERE clause to filter rows where Age > 15. The statement is:
SELECT Name, Class FROM Students WHERE Age > 15;

解答：所需 SQL 语句使用 SELECT 指定列 Name 和 Class，FROM 子句指明表 Students，WHERE 子句筛选条件 Age > 15。语句为：SELECT Name, Class FROM Students WHERE Age > 15;

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IGCSE Computer Science: Detailed Explanation of Typical Exam Questions | IGCSE 计算机：典型例题详解