Year 13 OCR Computer Science Mock Paper Walkthrough | A2 OCR 计算机科学模拟卷精讲

📚 Year 13 OCR Computer Science Mock Paper Walkthrough | A2 OCR 计算机科学模拟卷精讲

This article provides a detailed walkthrough of a mock exam paper for Year 13 OCR A Level Computer Science (H446). Each question is broken down with clear explanations and model answers, helping you consolidate key concepts across the specification and build confidence for the unit test.

本文详细解析一套 A2 OCR A Level 计算机科学（H446）模拟试卷。每道题目都配有清晰的讲解和模型答案，帮助你巩固考纲中的关键概念，为单元测试增强信心。

1. Processor Architecture: CISC vs RISC and Pipelining | 处理器架构：CISC 与 RISC 及流水线

CISC (Complex Instruction Set Computer) architectures, such as x86, feature a rich instruction set with variable-length instructions capable of performing multi‑step operations in hardware or microcode. They typically have fewer registers and rely on complex addressing modes.

CISC（复杂指令集计算机）架构（如 x86）拥有丰富的指令集，指令长度可变，可通过硬件或微码执行多步操作。它们通常寄存器数量较少，并依赖复杂的寻址模式。

RISC (Reduced Instruction Set Computer) designs like ARM use a small, highly optimised set of fixed‑length instructions. Only load and store operations access memory, while arithmetic instructions operate on registers. This allows hardwired control units, pipelining efficiency, and a large general‑purpose register file.

RISC（精简指令集计算机）设计（如 ARM）采用小型、高度优化的定长指令集。只有加载和存储指令访问内存，算术指令在寄存器上操作。这允许使用硬连线控制单元、高效的流水线以及大量通用寄存器组。

Pipelining overlaps the fetch, decode, execute, and write‑back stages of multiple instructions. However, hazards can stall the pipeline. A structural hazard arises when two instructions require the same hardware resource simultaneously. A data hazard occurs when an instruction depends on the result of a previous unfinished instruction, e.g., RAW (Read After Write). A control hazard is caused by branch instructions that flush the pipeline.

流水线技术将多条指令的取指、译码、执行和写回阶段重叠。然而，冒险可能导致流水线停顿。结构冒险发生在两条指令同时需要同一硬件资源时。数据冒险发生在指令依赖于前一条未完成指令的结果时，如 RAW（读后写）。控制冒险由分支指令引起，会清空流水线。

Data hazards can be mitigated by operand forwarding, where the ALU result is directly routed back to the next instruction’s input without waiting for the register write.

数据冒险可通过操作数前递（operand forwarding）来缓解，即将 ALU 结果直接路由到下一条指令的输入，而不必等待寄存器写回。

2. Data Structures: Stacks and Queues in Practice | 数据结构：栈和队列的实践

Given the operations Push 5, Push 8, Pop, Push 2, Pop, Push 4 on a stack, trace the final state. After Push 5 the stack is [5]; Push 8 gives [5,8]; Pop returns 8 and leaves [5]; Push 2 yields [5,2]; Pop returns 2 leaving [5]; Push 4 results in [5,4] with 4 at the top. This illustrates the LIFO (Last‑In‑First‑Out) principle.

给定栈操作：Push 5, Push 8, Pop, Push 2, Pop, Push 4，跟踪最终状态。Push 5 后栈为 [5]；Push 8 得 [5,8]；Pop 返回 8 留下 [5]；Push 2 得 [5,2]；Pop 返回 2 留下 [5]；Push 4 结果为 [5,4]，4 在栈顶。这体现了后进先出（LIFO）原则。

Stacks are widely used for subroutine call returns, interrupt handling, and syntax parsing. Queues implement FIFO (First‑In‑First‑Out) and are essential in print spooling, keyboard buffers, and breadth‑first search. A circular queue prevents wasted space by reusing vacated slots.

栈广泛用于子程序调用返回、中断处理和语法解析。队列实现先进先出（FIFO）原则，是打印假脱机、键盘缓冲和宽度优先搜索的基础。循环队列通过重用已释放空间来避免空间浪费。

3. Boolean Algebra Simplification | 布尔代数化简

Simplify the expression F = A·B·C + A·B·C’ + A·B’·C. First apply the distribution law: A·B·C + A·B·C’ = A·B·(C + C’) = A·B, because C + C’ = 1. The expression becomes F = A·B + A·B’·C.

化简表达式 F = A·B·C + A·B·C’ + A·B’·C。首先应用分配律：A·B·C + A·B·C’ = A·B·(C + C’) = A·B，因为 C + C’ = 1。表达式变为 F = A·B + A·B’·C。

Now factor A: F = A·(B + B’·C). Using the absorption theorem B + B’·C = B + C, we obtain the minimal form.

现在提取公因子 A：F = A·(B + B’·C)。利用吸收律 B + B’·C = B + C，得到最简式。

F = A·B + A·C

A Karnaugh map confirmation groups minterms 111, 110, and 101 to produce the two prime implicants A·B and A·C.

卡诺图验证：将小项 111、110 和 101 圈组，得到两个质蕴含项 A·B 和 A·C。

4. Floating Point Binary Representation | 浮点二进制表示

A 12‑bit floating point format

Published by TutorHao | Year 13 Computer Science Revision Series | aleveler.com

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