📚 Year 13 CIE Computer Science Unit Test Mock Paper Analysis | Year 13 CIE 计算机 单元测试模拟卷解析
This article walks through a complete mock unit test written for Year 13 CIE Computer Science students. Every question maps to a named syllabus area, and each answer is broken into the working that actually earns marks, so you can see exactly where candidates lose points and how to avoid it.
本文完整解析一套为 Year 13 CIE 计算机科学学生编写的单元测试模拟卷。每道题都对应考纲中的明确知识板块，每个答案都拆解到真正可以拿分的步骤，让你看清考生通常在哪里丢分，以及如何避免。
1. Number Systems and Data Representation | 数制与数据表示题目解析
Question 1 (4 marks). Convert the denary number 86 into 8-bit binary and hexadecimal. Then represent -86 using 8-bit two's complement, and state the full range of signed integers that can be stored in 8 bits.
第1题（4分）。将十进制数 86 转换为 8 位二进制与十六进制；然后用 8 位二进制补码表示 -86，并写出 8 位可存储的有符号整数范围。
2 = 2 remainder 1; 2 ÷ 2 = 1 remainder 0; 1 ÷ 2 = 0 remainder 1. Reading the remainders upwards gives 1010110, and padding to 8 bits with a leading zero gives 01010110.
方法：将 86 连续除以 2 并记录余数。86 ÷ 2 = 43 余 0；43 ÷ 2 = 21 余 1；21 ÷ 2 = 10 余 1；10 ÷ 2 = 5 余 0；5 ÷ 2 = 2 余 1；2 ÷ 2 = 1 余 0；1 ÷ 2 = 0 余 1。余数自底向上读取得到 1010110，前补一个 0 补齐 8 位即 01010110。
lip every bit to 10101001, then add 1 to get 10101010, which is 0xAA. The leading 1 confirms a negative value, and there is no separate minus sign stored in the byte.
十六进制只是把同一串比特按 4 位重新分组：0101 = 5，0110 = 6，因此答案是 0x56，写作 56₁₆。求 -86 时，从 +86 = 01010110 出发，按位取反得 10101001，再加 1 得 10101010，即 0xAA。最高位的 1 表明这是一个负数，字节中并不单独存储负号。
The range of 8-bit signed two's complement is -2⁷ to 2⁷ - 1, that is -128 to +127. The most negative pattern 10000000 has no positive counterpart, which is exactly why the range is asymmetric, and why adding two large positive numbers can flip the sign bit and produce overflow.
8 位有符号补码的范围是 -2⁷ 到 2⁷ - 1，即 -128 至 +127。最负的位模式 10000000 没有对应的正数，这正是范围不对称的原因，也正是两个较大的正数相加可能翻转符号位、产生溢出的原因。
| Step | Working | Mark |
| Binary | 01010110 | 1 |
| Hexadecimal | 56₁₆ | 1 |
| Two's complement | 10101010 | 1 |
| Range | -128 to +127 | 1 |
2. Boolean Algebra and Logic Gates | 布尔代数与逻辑门题目解析
Question 2 (4 marks). Simplify the Boolean expression F = ¬(A ∧ B) ∨ ¬(¬A ∨ B), then draw the logic circuit that implements the simplified expression.
第2题（4分）。化简布尔表达式 F = ¬(A ∧ B) ∨ ¬(¬A ∨ B)，并画出实现化简结果的逻辑电路。
Step one applies De Morgan's law to each negated bracket. ¬(A ∧ B) becomes ¬A ∨ ¬B. ¬(¬A ∨ B) becomes ¬(¬A) ∧ ¬B, and double negation removes the inner NOT, leaving A ∧ ¬B. Substituting both results gives F = ¬A ∨ ¬B ∨ (A ∧ ¬B).
第一步对每个取反的括号使用德摩根定律。¬(A ∧ B) 变为 ¬A ∨ ¬B。¬(¬A ∨ B) 变为 ¬(¬A) ∧ ¬B，双重否定消去内层非运算，剩下 A ∧ ¬B。代入两个结果得 F = ¬A ∨ ¬B ∨ (A ∧ ¬B)。
Step two applies the absorption law X ∨ (Y ∧ X) = X with X = ¬B. The term A ∧ ¬B is therefore absorbed by ¬B, and the whole expression collapses to F = ¬A ∨ ¬B, which by De Morgan is simply ¬(A ∧ B). The entire circuit is one NAND gate.
第二步使用吸收律 X ∨ (Y ∧ X) = X，此处 X = ¬B。因此 A ∧ ¬B 被 ¬B 吸收，整个表达式化简为 F = ¬A ∨ ¬B，按德摩根定律它就是 ¬(A ∧ B)。整个电路只需要一个与非门。
F = ¬(A ∧ B) ∨ ¬(¬A ∨ B) = ¬A ∨ ¬B = ¬(A ∧ B)
e usually awarded separately for applying De Morgan correctly, for the absorption step, and for a circuit diagram that matches the final expression rather than the original one.
真值表可以验证结果：输入对 00、01、10 时 F = 1，只有 11 时 F = 0。在考试中，分数通常分别给到正确应用德摩根定律、吸收律化简步骤，以及画出与最终表达式（而非原表达式）一致的电路图。
| A | B | ¬A ∨ ¬B | F |
| 0 | 0 | 1 | 1 |
| 0 | 1 | 1 | 1 |
| 1 | 0 | 1 | 1 |
| 1 | 1 | 0 | 0 |
Related exam favourites include the half adder, where Sum = A ⊕ B and Carry = A ∧ sing; a bare answer with no working often scores zero even when the final line is correct.
相关的高频考点还包括半加器，其中 Sum = A ⊕ B，Carry = A ∧ B，以及用卡诺图化简三变量或四变量表达式。答题时务必写明所用的定律；只有结论没有过程，即使最终一行正确也常常得零分。
3. Processor Fundamentals and the Fetch-Decode-Execute Cycle | 处理器基础与取指—译码—执行周期
Question 3 (6 marks). Describe what happens during the fetch stage of the fetch-decode-execute cycle. Explain the role of the program counter, the memory address register and the memory data register.
第3题（6分）。描述取指—译码—执行周期中取指阶段发生的过程，并解释程序计数器、存储器地址寄存器和存储器数据寄存器的作用。
During fetch, the address held in the program counter is copied to the memory address register, and that address is placed on the address bus. The contents of that memory location, which is the next instruction, travel along the data bus into the memory data register, and from there into the current instruction register. The program counter is then incremented so that it points to the next instruction in sequence.
取指阶段，程序计数器中保存的地址被复制到存储器地址寄存器，该地址经地址总线发送出去。该存储单元的内容（即下一条指令）经数据总线进入存储器数据寄存器，再送入当前指令寄存器。随后程序计数器自增，指向顺序中的下一条指令。
The program counter holds the address of the next instruction to be fetched. The memory address register holds the address currently being accessed in memory, and it is connected to the address bus. The memory data register acts as a buffer, holding the data or instruction that has just been read from, or is about to be written to, memory.
程序计数器保存将要取出的下一条指令的地址。存储器地址寄存器保存当前正在访问的存储地址，并与地址总线相连。存储器数据寄存器起缓冲作用，保存刚刚从内存读出、或即将写入内存的数据或指令。
In decode, the control unit splits the instruction into opcode and operand, and works out which operation is required. In execute, that operation is carried out, possibly using the arithmetic logic unit and the accumulator, with results written back to registers or memory. Modern processors overlap these stages using pipelining, which raises throughput because several instructions are at different stages at the same time.
译码阶段，控制单元把指令拆分为操作码与操作数，判断需要执行什么操作。执行阶段则真正完成该操作，可能用到算术逻辑单元和累加器，结果写回寄存器或内存。现代处理器用流水线技术重叠这些阶段，由于多条指令同时处于不同阶段，吞吐量得以提高。
Interrupts are also examined here. An interrupt is a signal from hardware or software that temporarily halts the current process; the processor saves the contents of registers and the program counter onto a stack, loads the address of the interrupt service routine from the interrupt vector table, services the interrupt, then restores the saved state. A question that asks you to compare a von Neumann machine with a Harvard machine simply wants you to explain that the former shares one memory for data and instructions, while the latter keeps them separate.
中断也是本部分的考点。中断是来自硬件或软件的信号，会暂时中止当前进程；处理器把寄存器和程序计数器的内容保存到栈中，从中断向量表取出中断服务程序的地址，处理完中断后再恢复已保存的状态。若题目要求比较冯·诺依曼机与哈佛机，只需说明前者用同一存储器存放数据与指令，后者则把两者分开存储。
4. Assembly Language and Addressing Modes | 汇编语言与寻址方式
Question 4 (5 marks). State what is meant by an addressing mode. Describe immediate, direct and indirect addressing, and give one situation in which indexed addressing is useful.
第4题（5分）。说明什么是寻址方式；描述立即寻址、直接寻址与间接寻址，并给出索引寻址适用的一个场景。
An addressing mode is the way in which the operand of an instruction specifies where the data to be operated on is held. The same mnemonic can therefore work with a number typed into the instruction, a value in a named memory location, or a value whose address is itself stored elsewhere.
寻址方式是指指令的操作数如何指明被处理数据的存放位置。因此同一条助记符指令，既可以对直接写在指令中的数值操作，也可以对某个具名存储单元的值操作，还可以对地址本身存放在别处的值操作。
In immediate addressing, the operand is the actual value, for example LDA #7 loads the number 7 into the accumulator. In direct addressing, the operand is the address of the data, for example LDA 300 loads whatever is stored at address 300. In indirect addressing, the operand gives the address of a location that itself holds the address of the data, which is written LDA [300] and is essential for pointers and for jumping to a return address held on the stack.
立即寻址中，操作数就是实际数值，例如 LDA #7 把数字 7 装入累加器。直接寻址中，操作数就是数据的地址，例如 LDA 300 装入地址 300 处存放的内容。间接寻址中，操作数给出的是一个存储单元的地址，而该单元里存放的才是数据的地址，写作 LDA [300]，它对于指针以及跳转到栈中保存的返回地址是必不可少的。
Indexed addressing adds the contents of an index register to a stated base address, so the same instruction can step through consecutive items. It is useful for traversing arrays and for processing records in a file, where the offset changes by a fixed amount each time. Assembly language itself is a low-level language using mnemonics, one instruction per machine-code instruction, translated by an assembler, while a compiler translates a high-level language as a whole and an interpreter translates and executes line by line.
索引寻址把变址寄存器的内容加到指定的基地址上，因此同一条指令可以逐个访问连续的数据项。它适用于遍历数组，以及在文件中处理记录，因为每次偏移量都按固定大小变化。汇编语言本身是一种使用助记符的低级语言，一条指令对应一条机器码指令，由汇编器翻译；而编译器整体翻译高级语言，解释器则逐行翻译并执行。
5. Data Structures: Stacks, Queues, Linked Lists and Trees | 数据结构：栈、队列、链表与树
Question 5 (6 marks). Explain the difference between a stack and a queue, and state one practical use of each. Describe how a linked list differs from an array, and give one advantage of each.
第5题（6分）。解释栈与队列的区别并各给出一个实际用途；说明链表与数组的不同，并各给出一个优点。
A stack is a last in, first out structure: the most recently pushed item is the first one popped. Items are added and removed at the same end, called the top. A queue is first in, first out: items join at the rear and leave from the front. Push and pop operate on a stack, while enqueue and dequeue operate on a queue.
栈是后进先出结构：最近压入的元素最先弹出，元素的添加与删除都发生在同一端，称为栈顶。队列是先进先出：元素从队尾加入，从队首离开。栈使用 push 与 pop 操作，队列使用 enqueue 与 dequeue 操作。
Typical uses: a stack stores return addresses during nested subroutine calls, and reverses data. A queue buffers print jobs so documents are printed in the order they were submitted. Both can be implemented using an array with a pointer, or using a linked list, which removes any fixed size limit.
典型用途：栈在嵌套子程序调用中保存返回地址，也可用于数据反转。队列用于缓冲打印任务，使文档按提交顺序打印。两者都可用带指针的数组实现，也可用链表实现，后者去掉了固定容量限制。
An array stores its elements in contiguous memory locations, so any element can be accessed directly in constant time using its index. A linked list stores nodes that may be scattered in memory, each holding data plus a pointer to the next node, so items can be inserted or deleted without shifting the rest of the structure, at the cost of no direct random access and extra memory for the pointers.
数组把元素存放在连续的内存单元中，因此可用下标在常数时间内直接访问任意元素。链表存放的结点在内存中可以分散，每个结点保存数据以及指向下一个结点的指针，因此插入或删除元素无需移动其余数据，代价是无法直接随机访问，并且要额外占用指针空间。
A binary search tree keeps smaller values in the left subtree and larger values in the right, giving efficient searching when the tree is balanced; a traversal such as in-order visits nodes in sorted order. Hash tables give near constant-time lookup but need a collision policy, and their worst case degrades to linear search.
二叉搜索树把较小的值放在左子树、较大的值放在右子树，在树平衡时查找效率很高；中序遍历等遍历方法可以按排序顺序访问结点。哈希表提供接近常数时间的查找，但需要冲突处理策略，最坏情况下会退化为线性查找。
6. Networks, Protocols and the TCP/IP Stack | 网络、协议与 TCP/IP 协议栈
Question 6 (6 marks). Name the four layers of the TCP/IP stack and describe the role of each. Explain the difference between TCP and UDP, and give one situation in which UDP is preferable.
第6题（6分）。写出 TCP/IP 协议栈的四个层次并描述各自作用；说明 TCP 与 UDP 的区别，并给出适合使用 UDP 的一个场景。
The four layers are application, transport, internet and link. The application layer provides services to the user and to programs, and includes HTTP, HTTPS, FTP, SMTP, IMAP and DNS. The transport layer splits data into segments, handles end-to-end delivery, and contains TCP and UDP. The internet layer handles logical addressing and routing of packets, and contains IP, with ICMP for diagnostics. The link layer deals with physical transmission and hardware addressing such as MAC addresses, and includes Ethernet and Wi-Fi.
四层分别是应用层、传输层、网际层和链路层。应用层为用户和程序提供服务，包含 HTTP、HTTPS、FTP、SMTP、IM
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