GCSE WJEC Computer Science: Typical Example Questions Explained in Detail | GCSE WJEC 计算机科学：典型例题详解

📚 GCSE WJEC Computer Science: Typical Example Questions Explained in Detail | GCSE WJEC 计算机科学：典型例题详解

WJEC GCSE Computer Science challenges students with a mix of theoretical understanding and practical computational thinking. This article walks through typical exam-style questions, providing step-by-step solutions and explanations to help you master core concepts.

WJEC 的 GCSE 计算机科学考试要求学生兼顾理论理解与实际计算思维。本文逐一讲解典型考题，提供分步详解与解析，帮助你掌握核心知识。

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1. Binary Arithmetic | 二进制算术

In Unit 1, you must be confident in binary addition and two’s complement subtraction. A common exam task involves adding 8-bit binary numbers and detecting overflow.

在 Unit 1 中，你必须熟练掌握二进制加法与二进制补码减法。考试中常见的任务是对 8 位二进制数进行加法运算并判断是否溢出。

Example 1: Add the 8-bit numbers 01001101₂ and 00111010₂. Show carry bits and state whether an overflow occurs.

例题 1：将 8 位二进制数 01001101₂ 与 00111010₂ 相加。写出进位过程，并说明是否发生溢出。

Solution: Start from the least significant bit (right). 1+0 = 1 (carry 0). Next: 0+1 = 1 (carry 0). 1+0 = 1 (carry 0). 1+1 = 0 carry 1. 0+1+carry 1 = 0 carry 1. 0+1+carry 1 = 0 carry 1. 1+0+carry 1 = 0 carry 1. Finally, 0+0+carry 1 = 1. The result is 10000111₂. There is no overflow because the sum of two positive numbers (01001101 is +77, 00111010 is +58) yields 135, which fits within -128 to 127 in 8-bit two’s complement? Wait: these are unsigned or two’s complement? The question implies unsigned or that the range check uses carry out of MSB. Actually, if treated as unsigned, result 135 is fine. In WJEC, they often ask to check overflow by looking at carry into and out of the most significant bit: here carry into MSB is 0 and carry out is 0 – no overflow.

解答：从最低有效位（最右侧）开始。1+0 = 1（进位 0）。下一列：0+1 = 1（进位 0）。1+0 = 1（进位 0）。1+1 = 0 进位 1。0+1+进位 1 = 0 进位 1。0+1+进位 1 = 0 进位 1。1+0+进位 1 = 0 进位 1。最后 0+0+进位 1 = 1。结果为 10000111₂。判断溢出：若看作有符号数，最高位的进位输入为 0，进位输出为 0，两者相同，无溢出。若看作无符号整数，135 未超过 255，同样无溢出。

Example 2: Represent the decimal number -18 in 8-bit two’s complement.

例题 2：用 8 位二进制补码表示十进制数 -18。

Solution: Write +18 in binary: 00010010₂. Invert bits: 11101101₂. Add 1: 11101110₂. This is the two’s complement representation of -18.

解答：先写出 +18 的二进制形式：00010010₂。将所有位取反：11101101₂。再加 1，得到 11101110₂。这就是 -18 的 8 位二进制补码表示。

To subtract using two’s complement, simply add the complement. For example, 45 – 18 becomes 45 + (-18). This method is regularly tested.

用补码做减法时，只需将减数转为补码后相加。例如 45 – 18 即为 45 + (-18)。这种方法在考试中时常出现。

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

WJEC expects you to analyse combinational logic circuits. A typical question gives a diagram described in words and asks for the Boolean expression and truth table.

WJEC 要求考生能分析组合逻辑电路。典型题目会以文字描述电路，要求写出布尔表达式并填写真值表。

Example: A circuit has inputs A, B and C. Gate 1 is an AND gate receiving A and B. Gate 2 is an OR gate receiving the output of Gate 1 together with C. Write the expression for the overall output Q, and complete its truth table.

例题：某电路有输入 A、B、C。门 1 是与门，输入为 A 和 B；门 2 是或门，其输入为门 1 的输出以及 C。写出总输出 Q 的表达式，并完成真值表。

Expression: Q = (A AND B) OR C, often written as Q = (A.B) + C.

表达式：Q = (A AND B) OR C，常写作 Q = (A · B) + C。

Truth table:

A	B	C	A AND B	Q
0	0	0	0	0
0	0	1	0	1
0	1	0	0	0
0	1	1	0	1
1	0	0	0	0
1	0	1	0	1
1	1	0	1	1
1	1	1	1	1

Always label intermediate columns to show working. This helps secure method marks even if the final column has an error.

务必为中间列加上标签，展示分析过程。即便最终列出现错误，这也有助于获得方法分。

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3. Trace Tables in Pseudocode | 伪代码跟踪表

Tracking variable values through pseudocode is a key skill for Unit 2. You are often given a loop and asked to complete a trace table.

通过伪代码追踪变量值是 Unit 2 的关键技能。考试常给出一个循环，要求你完成跟踪表。

Example: Study the following pseudocode and complete the trace table for total and i.

例题：阅读以下伪代码，并完成变量 total 和 i 的跟踪表。

total ← 0
FOR i ← 1 TO 5
  total ← total + i * 2
ENDFOR
OUTPUT total

Trace table – fill in the values after each iteration:

Iteration	i	total
initial	–	0
1	1	2
2	2	6
3	3	12
4	4	20
5	5	30

The final output is 30. Trace tables test careful sequencing and help diagnose logic errors in your own code.

最终输出为 30。跟踪表考查严谨的顺序思维，也有助于在你自己编写的代码中诊断逻辑错误。

When the loop uses condition-controlled logic (WHILE, REPEAT), be extra careful to note when the condition fails so that the loop exits correctly.

当循环采用条件控制（WHILE、REPEAT）时，要特别注意条件为假的那一刻，以便循环能正确退出。

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4. Sorting Algorithms: Bubble Sort Example | 排序算法：冒泡排序例题

Bubble sort is a standard GCSE algorithm. You may be asked to show the state of the array after each pass or to identify how many passes are needed.

冒泡排序是 GCSE 的标准算法。你可能需要展示每趟排序后的数组状态，或判断总共需要多少趟排序。

Example: Apply the bubble sort algorithm to the array [5, 2, 8, 1, 3]. Show the array after each complete pass and indicate when no swaps occur.

例题：对数组 [5, 2, 8, 1, 3] 应用冒泡排序算法。展示每完整一趟之后的数组，并标明何时没有发生交换。

Pass 1: Compare and swap if needed. 5>2 → swap → [2,5,8,1,3]; 5<8 → no swap; 8>1 → swap → [2,5,1,8,3]; 8>3 → swap → [2,5,1,3,8]. End of pass 1.

第 1 趟：比较并依需交换。5>2 → 交换 → [2,5,8,1,3]；5<8 → 不交换；8>1 → 交换 → [2,5,1,8,3]；8>3 → 交换 → [2,5,1,3,8]。第 1 趟结束。

Pass 2: [2,5,1,3,8] → 2<5 no swap; 5>1 swap → [2,1,5,3,8]; 5>3 swap → [2,1,3,5,8]; 5<8 no swap → [2,1,3,5,8]. Swaps occurred.

第 2 趟：[2,5,1,3,8] → 2<5 不交换；5>1 交换 → [2,1,5,3,8]；5>3 交换 → [2,1,3,5,8]；5<8 不交换 → [2,1,3,5,8]。发生了交换。

Pass 3: [2,1,3,5,8] → 2>1 swap → [1,2,3,5,8]; rest in order, no swaps. Pass 4 would have no swaps, so algorithm terminates early. Sorted after 3 passes.

第 3 趟：[

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