📚 IGCSE CIE Computer Science Common Mistake Questions Explained | IGCSE CIE 计算机：易错题精讲
In IGCSE CIE Computer Science, many students lose marks not because they don’t understand the theory, but because they fall into predictable traps set by examiners. Misreading units, confusing similar algorithms, mishandling binary arithmetic, or making small syntax errors in pseudocode and SQL can turn a confident answer into a wrong one. This article collects the most common mistake questions reported by teachers and examiners, breaks down why they cause trouble, and shows you exactly how to avoid them. Each section pairs a typical error with a clear, step‑by‑step correction, so you can sharpen your exam technique and secure the grades you deserve.
在 IGCSE CIE 计算机科学考试中，许多学生丢分并非因为不懂理论，而是掉入了考官精心设计的常见陷阱。读错单位、混淆相似算法、二进制算术操作不当，或者伪代码与 SQL 中的小语法错误，都能让本来信心满满的答案变成错误答案。本文汇集了教师和考官反馈中最常见的易错题，分析出错原因，并给出精准的避错方法。每一节将典型错误与清晰的分步正解配对，帮助你强化应试技巧，稳稳拿到应得的分数。
1. Hexadecimal and Binary Conversion Traps | 十六进制与二进制转换陷阱
A very common mistake is treating hexadecimal digits as two separate decimal digits when converting to binary. For example, converting the denary number 200 into hexadecimal: 200 encodes as C8, but many students write ’12’ and ‘8’ then convert 12 to 1100 and 8 to 1000, which actually does give C8 correctly. The error appears when converting a hex digit like A (which is 10) directly to binary: students sometimes write 1010 as ‘1’ and ‘0’ separately, or they forget to pad to 4 bits for each hex digit.
一个极为常见的错误是在将十六进制转换为二进制时，把每个十六进制位当作两个独立的十进制数字来处理。例如，将十进制数 200 转换为十六进制：200 编码为 C8，但很多学生会写出 ’12’ 和 ‘8’，然后把 12 转换为 1100、8 转换为 1000，这一步其实碰巧得到了正确的 C8。然而，当转换像 A（即 10）这样的十六进制位时，问题就出现了：学生有时会把 1010 拆成 ‘1’ 和 ‘0’ 分别处理，或者忘记每个十六进制位必须补齐 4 位二进制。
Correct approach: Each hex digit must be replaced by exactly 4 binary bits, using leading zeros where needed. For 2F: 2 → 0010, F → 1111, so 00101111. The most frequent exam pitfall: converting denary 20 to hex. 20 = 14 in hex, but students incorrectly write 14 as 1 and 4, then 1→0001 and 4→0100, obtaining 00010100, which is actually 20 in binary because 00010100 is 14 in hex? Wait: 14 in hex is 20 denary. That’s correct, but the error is when they treat 14 as two denary digits 1 and 4, which coincidentally gives right binary for 20 denary? Let’s check: 14 hex = 1×16+4 = 20 denary. If you convert 1 as 0001 and 4 as 0100, you get 00010100 binary, which is indeed 20 denary. So it’s correct here. The real danger: converting denary 26 to hex gives 1A. Students might write 1 and A, then 1→0001, A→1010 → 00011010, which is right. The mistake usually arises when they forget that hex A–F are numbers, and they try to convert ‘A’ as two digits or use ASCII. So always treat A=10, B=11, etc., and map to 4‑bit binary.
正确方法：每个十六进制位必须替换为恰好 4 位二进制，必要时前补零。例如 2F：2 → 0010，F → 1111，得到 00101111。考试中最容易出错的情况：忘记十六进制 A–F 本身就是数值，试图将其当作两个字符处理，或者不补齐 4 位。记住 A=1010，B=1011，… F=1111。并始终确保每组 4 位。
2. Binary Addition Overflow Errors | 二进制加法溢出错误
When adding two 8‑bit binary numbers, students often forget to check for overflow, or they mistakenly think a carry into the 9th bit always means an overflow. Overflow only occurs when the result exceeds the range representable by the given number of bits and the two operands have the same sign (in two’s complement). For an 8‑bit unsigned addition, a 9th carry simply means the result is larger than 255, which is overflow. But for two’s complement, a carry out of the most significant bit alone does not indicate overflow; you must compare the carry into the MSB with the carry out of the MSB. If they differ, overflow occurs.
在对两个 8 位二进制数进行加法时，学生经常忘记检查溢出，或者错误地认为只要向第 9 位产生了进位就一定是溢出。溢出仅在结果超出指定位数所表示的范围，并且两个操作数符号相同（在二进制补码中）时才发生。对于 8 位无符号加法，第 9 位进位意味着结果大于 255，即溢出。但在二进制补码中，仅最高有效位（MSB）的进位不足以判断溢出；必须比较进入 MSB 的进位与出 MSB 的进位。如果两者不同，则发生溢出。
Example: Add 01100010 (98) and 00111101 (61) in unsigned 8‑bit. 98+61=159, within 0–255, no overflow. But add 10000001 (129) and 10000010 (130), 129+130=259, requires 9 bits, overflow in unsigned. For two’s complement, 10000001 is –127, 10000010 is –126, sum is –253, which is outside –128 to 127 range, overflow. The CPU flags this; in exams, you must state ‘overflow occurs because the result is outside the representable range’. Students often just write ‘carry 1’ and stop, missing the explanation mark.
示例：将 01100010 (98) 和 00111101 (61) 进行 8 位无符号加法。98+61=159，在 0–255 范围内，无溢出。但将 10000001 (129) 和 10000010 (130) 相加，129+130=259，需要 9 位，无符号溢出。在二进制补码中，10000001 是 –127，10000010 是 –126，和为 –253，超出 –128 到 127 范围，溢出。考试中必须明确写出“溢出发生是因为结果超出了可表示范围”。许多学生只写“进位 1”就停下，丢了解释分。
3. Logic Gate Misinterpretations: NAND and NOR | 逻辑门误读：NAND 与 NOR
A persistent error is drawing the wrong truth table for NAND and NOR gates, or confusing them with AND/OR followed by a NOT bubble placed incorrectly. A NAND gate outputs 1 unless both inputs are 1. A NOR gate outputs 1 only when both inputs are 0. Students often recall ‘opposite of AND’ but then write the truth table as 0 only when both 1 – that’s correct, but they may incorrectly cascade gates without using brackets in Boolean expressions.
一个持久性的错误是为 NAND 和 NOR 门绘制错误的真值表，或者将其与 AND/OR 后加错非门气泡混淆。NAND 门除非两个输入都为 1，否则输出 1。NOR 门仅当两个输入都为 0 时才输出 1。学生通常记得“与 AND 相反”，但有时写出的真值表是仅当两个输入为 1 时输出 0——这其实正确。然而，在布尔表达式中级联门时，他们经常忽略括号，导致运算顺序错误。
Exam trap: Draw a circuit for (A AND B) OR (NOT C). Students might place a NOT gate on C, then AND A and B, then OR. But if the question asks for NAND equivalence, they get lost. Another common mistake: when drawing a logic gate from a truth table, they miss the pattern of a XOR gate and instead use a combination of AND/OR that becomes unnecessarily complex. Always simplify using Karnaugh maps if required, and double‑check the output column for each input combination.
考试陷阱：绘制 (A AND B) OR (NOT C) 的逻辑电路。学生可能会在 C 上放一个非门，然后 A 和 B 进行与运算，最后或。但如果题目要求用与非门实现，他们就迷失了。另一个常见错误：从真值表绘制逻辑门时，未能识别出异或门（XOR）的模式，反而使用了不必要的 AND/OR 复杂组合。如果要求，务必使用卡诺图化简，并仔细核对每种输入组合的输出列。
4. Storage Units: Kibibyte vs Kilobyte Confusion | 存储单位：Kibibyte 与 Kilobyte 混淆
IGCSE CIE often expects you to know that 1 kilobyte (kB) = 1000 bytes, while 1 kibibyte (KiB) = 1024 bytes, following IEC standards. Many students use 1024 for everything, which is acceptable in older syllabi but CIE has moved to clarify the distinction. A typical error question: ‘A file size is 2 MiB. What is its size in bytes?’ Students multiply 2 by 1000² instead of 1024². The correct answer is 2 × 1024 × 1024 = 2,097,152 bytes. When converting between units, always check the prefix: kibi (Ki), mebi (Mi), gibi (Gi) use 1024; kilo (k), mega (M), giga (G) use 1000. Questions about data transfer rates (kbps vs KiB/s) also trap students.
IGCSE CIE 通常期望学生了解 1 千字节 (kB) = 1000 字节，而 1 kibibyte (KiB) = 1024 字节，遵循 IEC 标准。许多学生对所有情况都使用 1024，这在旧大纲中可接受，但 CIE 已转向明确区分。典型错误题：“文件大小为 2 MiB，其字节大小是多少？”学生用 2 × 1000² 而不是 1024²。正确答案是 2 × 1024 × 1024 = 2,097,152 字节。在单位转换时，务必检查前缀：kibi (Ki)、mebi (Mi)、gibi (Gi) 使用 1024；kilo (k)、mega (M)、giga (G) 使用 1000。有关数据传输率的问题 (kbps vs KiB/s) 也常设陷阱。
Tip: When the exam asks for ‘kilobyte’ and does not specify ‘kibibyte’, assume 1000, but be careful: if the question is about RAM or memory sizes, traditionally 1024 was used. CIE questions now explicitly write KiB, MiB to avoid ambiguity. If you see KB (upper case K), it usually means 1024. Always read the unit symbol carefully and apply the correct power: 10³ for k, 2¹⁰ for Ki.
提示：当考试要求“千字节”且未指明“kibibyte”时，假设为 1000，但需谨慎：如果问题涉及 RAM 或存储器大小，传统上使用 1024。现在 CIE 的题目明确写为 KiB、MiB 以避免歧义。如果你看到 KB（大写 K），它通常表示 1024。始终仔细阅读单位符号并应用正确的乘方：k 为 10³，Ki 为 2¹⁰。
5. Pseudocode Loop Execution Counts | 伪代码循环执行次数
A classic pitfall is misjudging how many times a FOR loop runs. In CIE pseudocode, FOR i ← 1 TO 10 executes 10 times, with i taking values 1 through 10. But students often think it runs 11 times (including a phantom 0 or 11). Another error: FOR i ← 10 TO 1 STEP –1, some assume it won’t run at all because the start is larger than the end, but with a negative step it correctly runs 10 times. The boundary condition FOR i ← 1 TO n is n times. When arrays are involved, off‑by‑one errors cause indexes out of bounds.
经典陷阱：错误判断 FOR 循环运行的次数。在 CIE 伪代码中，FOR i ← 1 TO 10 执行 10 次，i 取值 1 到 10。但学生经常以为它运行 11 次（包括一个虚幻的 0 或 11）。另一个错误：FOR i ← 10 TO 1 STEP –1，有些人认为它不会运行，因为开始大于结束，但由于步长为负，它正确运行 10 次。边界条件 FOR i ← 1 TO n 运行 n 次。当涉及数组时，差一错误会导致下标越界。
Example of error: A question asks ‘How many times does the loop PRINT i? FOR i ← 0 TO 9 OUTPUT i’. The answer is 10 (0–9). Students write 9. To avoid this, count inclusive endpoints: number of iterations = (end – start)/step + 1, assuming step divides evenly. For nested loops, multiply the counts.
错误示例：题目问“循环 PRINT i 执行了多少次？FOR i ← 0 TO 9 OUTPUT i”。答案是 10 (0–9)。学生写 9。为避免出错，闭区间端点计数：迭代次数 = (结束值 – 起始值)/步长 + 1，假设步长整除。对于嵌套循环，将各层计数相乘。
6. Bubble Sort vs Insertion Sort Confusion | 冒泡排序与插入排序混淆
Students often confuse the mechanics of bubble sort and insertion sort when tracing. In bubble sort, adjacent pairs are swapped if out of order, and after each pass the largest unsorted element ‘bubbles’ to its correct position at the end. In insertion sort, each element is taken from the unsorted part and inserted into its correct position in the sorted portion, shifting elements right to make space. A common exam question shows a partially sorted list and asks which algorithm produced it, or asks for the number of comparisons after a specific pass. Candidate mistakes: counting passes as comparisons, or thinking insertion sort compares only adjacent elements.
在跟踪执行时，学生经常混淆冒泡排序和插入排序的机制。在冒泡排序中，如果相邻元素顺序不对就交换，每趟之后最大的未排序元素“冒泡”到末尾的正确位置。在插入排序中，从未排序部分取出每个元素，并将其插入到已排序部分的正确位置，右移元素以腾出空间。常见的考题展示一个部分排序的列表，询问是哪种算法产生的，或者要求计算特定趟后的比较次数。考生常见错误：将趟数计为比较次数，或者以为插入排序只比较相邻元素。
Correct tracing method: For bubble sort on [5,1,4,2,8]: Pass 1 compares (5,1) swap → [1,5,4,2,8]; (5,4) swap → [1,4,5,2,8]; (5,2) swap → [1,4,2,5,8]; (5,8) no swap. After pass 1, 8 is in final position. Insertion sort: start with [5]; next 1 < 5 → insert before 5 → [1,5]; next 4: compare 4 with 5, shift 5, insert 4 → [1,4,5]; next 2: compare 2 with 5, shift; compare 2 with 4, shift; compare 2 with 1, stop → [1,2,4,5]; etc. Knowing the distinction helps answer algorithms questions accurately.
正确跟踪方法：对于冒泡排序 [5,1,4,2,8]：第 1 趟比较 (5,1) 交换 → [1,5,4,2,8]; (5,4) 交换 → [1,4,5,2,8]; (5,2) 交换 → [1,4,2,5,8]; (5,8) 不交换。第 1 趟后，8 已在最终位置。插入排序：以 [5] 开始；下一个 1 < 5 → 插入到 5 之前 → [1,5]；下一个 4：4 与 5 比较，5 右移，插入 4 → [1,4,5]；下一个 2：2 与 5 比较，移位；2 与 4 比较，移位；2 与 1 比较，停止 → [1,2,4,5]；依此类推。理解这一区别有助于准确解答算法题。
7. Check Digit Calculation Mistakes | 校验位计算错误
ISBN‑13 and barcode check digit questions are common. A typical mistake is applying the wrong weight pattern or forgetting to subtract the remainder from 10 when finding the check digit. For ISBN‑13: alternating weights of 1 and 3, calculate sum, divide by 10, then check digit = 10 – remainder (if remainder 0, check digit is 0). Many students just calculate the sum and choose the remainder as check digit, which is incorrect. Also, they sometimes use 1‑based indexing incorrectly, assigning weight 1 to the first digit even though the pattern changes.
ISBN‑13 及条形码校验位问题十分常见。典型错误是使用错误的权重模式，或者在求校验位时忘记用 10 减去余数。对于 ISBN‑13：权重交替为 1 和 3，计算总和，除以 10，然后校验位 = 10 – 余数（如果余数为 0，校验位为 0）。许多学生只计算总和并选择余数作为校验位，这是错误的。此外，他们有时错误地使用基于 1 的索引，即使模式改变，也将权重 1 分配给第一个数字。
Example: Given first 12 digits of ISBN 978‑0‑13‑149642‑?, the last digit missing. Calculate: 9×1 + 7×3 + 8×1 + 0×3 + 1×1 + 3×3 + 1×1 + 4×3 + 9×1 + 6×3 + 4×1 + 2×3 = 9+21+8+0+1+9+1+12+9+18+4+6 = 98. 98 mod 10 = 8, so check digit = 10 – 8 = 2. Students who mistakenly use remainder 8 as check digit would get 8, which fails validation. Always verify your final check digit by re‑calculating the sum of all 13 digits (with check digit included) and ensuring it is divisible by 10.
示例：给定 ISBN 前 12 位 978‑0‑13‑149642‑?，最后一位缺失。计算：9×1 + 7×3 + 8×1 + 0×3 + 1×1 + 3×3 + 1×1 + 4×3 + 9×1 + 6×3 + 4×1 + 2×3 = 9+21+8+0+1+9+1+12+9+18+4+6 = 98。98 mod 10 = 8，因此校验位 = 10 – 8 = 2。错误地使用余数 8 作为校验位的学生会得到 8，验证不通过。务必通过重新计算所有 13 位数字（含校验位）的总和来验证，确认总和能被 10 整除。
8. Network Protocol Layer Mix‑ups | 网络协议分层混淆
When asked which layer a protocol belongs to, students frequently misplace HTTP and FTP. In the TCP/IP model used by CIE, Application layer includes HTTP, FTP, SMTP, IMAP. Transport layer includes TCP and UDP. Internet layer includes IP. Network Access layer includes Ethernet and Wi‑Fi. A common error is placing TCP in the Internet layer, or putting IP in the Transport layer. Remember the mnemonic ‘All Tomatoes Inside Nana’ (Application, Transport, Internet, Network) – but that can be misleading. Better to understand that TCP ensures reliable delivery, while IP handles addressing and routing.
当被问及某个协议属于哪一层时，学生经常将 HTTP 和 FTP 放错位置。在 CIE 采用的 TCP/IP 模型中，应用层包括 HTTP、FTP、SMTP、IMAP。传输层包括 TCP 和 UDP。互联网层包括 IP。网络接入层包括以太网和 Wi‑Fi。一个常见错误是将 TCP 放在互联网层，或者将 IP 放在传输层。记住助记符“All Tomatoes Inside Nana”（应用层、传输层、互联网层、网络层）可能有误导。更好的方式是理解 TCP 确保可靠传输，而 IP 处理寻址和路由。
Exam question style: ‘State the layer responsible for routing packets between networks.’ Answer: Internet layer. Students incorrectly say Transport. Also, when explaining the role of a switch vs a router, they mix layers: a switch operates at the Network Access layer (MAC addresses), a router at the Internet layer (IP addresses). Remember: L2 switch, L3 router.
考题风格：“陈述负责在网络之间路由数据包的层。”答案：互联网层。学生会错误地说传输层。此外，在解释交换机和路由器的角色时，他们混淆层次：交换机在网络接入层运行（MAC 地址），路由器在互联网层运行（IP 地址）。记牢：二层交换机、三层路由器。
9. SQL Query Syntax and Logic Errors | SQL 查询语法与逻辑错误
In practical theory papers, students write SQL statements incorrectly. Common mistakes: omitting the semicolon, forgetting quotation marks around string literals, using = instead of LIKE for pattern matching, and getting the WHERE clause logic wrong. For example, ‘SELECT * FROM Students WHERE Grade = ‘A’ OR Grade = ‘B” is correct, but many write ‘Grade = ‘A’ OR ‘B”, which evaluates wrong. Also, when using ORDER BY, they sometimes place DESC in the SELECT line. Another typical error: mixing column names and values without table prefix when using JOIN.
在理论笔试卷中，学生书写 SQL 语句时常出错。常见错误：遗漏分号、字符串文字周围忘记加引号、模式匹配时用 = 而不是 LIKE，以及 WHERE 子句逻辑错误。例如，’SELECT * FROM Students WHERE Grade = ‘A’ OR Grade = ‘B” 是正确的，但许多人写成 ‘Grade = ‘A’ OR ‘B”，这会错误求值。此外，使用 ORDER BY 时，他们有时会将 DESC 放在 SELECT 行。另一个典型错误：在使用 JOIN 时，列名和值未加表前缀导致混淆。
Correction: Always use single quotes for strings in CIE’s assumed SQL. Write AND, OR with complete conditions: condition1 AND condition2. For JOIN, specify table.column. When using aggregate functions like COUNT, GROUP BY is necessary if selecting non‑aggregated columns. Students often forget GROUP BY and get an error, or they include a non‑aggregated column in SELECT without it appearing in GROUP BY. Check your syntax systematically.
纠正：在 CIE 假定的 SQL 中，字符串始终使用单引号。使用完整的条件书写 AND、OR：condition1 AND condition2。对于 JOIN，指定 table.column。使用聚合函数如 COUNT 时，如果 SELECT 中包含非聚合列，则必须有 GROUP BY。学生经常忘记 GROUP BY 导致错误，或者在 SELECT 中包含非聚合列却未将其放入 GROUP BY。应系统检查语法。
10. Lossy vs Lossless Compression Misapplication | 有损与无损压缩的误用
Students know the definitions but fail to apply them to scenarios. They might say JPEG is lossy, but then state that text files can be compressed with JPEG – obviously wrong. A typical question: ‘Which compression method is suitable for a text document and why?’ Answer: lossless (e.g., ZIP, run‑length encoding) because every bit of data must be restored exactly; lossy would cause loss of text characters. Another error: thinking lossless always achieves higher compression ratios than lossy; in reality, lossy can achieve far greater compression where some quality loss is acceptable.
学生知道定义，但无法将其应用到实际场景。他们可能会说 JPEG 是有损的，但随后声称文本文件可以用 JPEG 压缩——显然错误。典型问题：“哪种压缩方法适用于文本文档，为什么？”答案：无损（例如 ZIP、游程编码），因为每一比特数据都必须精确恢复；有损会导致文本字符丢失。另一个错误：认为无损压缩的压缩率总是高于有损；现实中，在可接受一定质量损失的情况下，有损可以实现更大的压缩率。
Common exam request: ‘Explain why MP3 is used for audio rather than FLAC.’ Answer: MP3 is lossy, removes frequencies the human ear cannot hear, greatly reduces file size; FLAC is lossless, preserves all audio data, larger size. Choosing the right compression for a given data type is a key skill. Also, make sure you can describe how run‑length encoding works: replacing repeated consecutive characters with a code of character + count.
常见考题要求：“解释为什么音频使用 MP3 而不是 FLAC。”答案：MP3 是有损压缩，去除了人耳听不到的频率，大幅减小文件大小；FLAC 是无损压缩，保留所有音频数据，尺寸较大。为给定数据类型选择合适的压缩方式是关键技能。此外，确保你能描述游程编码的工作原理：用字符加计数替换连续重复的字符。
11. Symmetric vs Asymmetric Encryption Traps | 对称加密与非对称加密的陷阱
A frequent misconception is that symmetric encryption uses two different keys (it uses one shared key), and asymmetric uses one key (it uses a public/private pair). The exam will often ask: ‘Which type of encryption is used for sending a secure email?’ The answer involves asymmetric encryption: the sender encrypts the email with the recipient’s public key; the recipient decrypts with their private key. Students sometimes write symmetric because they think it’s faster, but for key exchange without prior shared secret, asymmetric is needed. Also, mixing up digital signatures: a digital signature uses the sender’s private key to encrypt a hash, providing non‑repudiation.
一个常见的误解是认为对称加密使用两个不同的密钥（实际上它使用一个共享密钥），而非对称加密使用一个密钥（它使用公钥/私钥对）。考试经常会问：“发送安全电子邮件使用哪种加密方式？”答案涉及非对称加密：发送方用收件人的公钥加密邮件；收件人用自己的私钥解密。学生有时会写对称加密，因为他们认为它更快，但在没有事先共享秘密进行密钥交换的情况下，需要使用非对称加密。另外，混淆数字签名：数字签名使用发送方的私钥加密哈希值，提供不可否认性。
Clarification: Symmetric: same key encrypts and decrypts, fast, used for bulk data (AES). Asymmetric: public key encrypts, private key decrypts; slow, used for key exchange and digital signatures (RSA). A combined system (hybrid) uses asymmetric to share a symmetric session key, then symmetric for data transfer. This appears in SSL/TLS. Understanding these roles prevents blunders.
澄清：对称加密：同一密钥加解密，速度快，用于大量数据（AES）。非对称加密：公钥加密，私钥解密；速度慢，用于密钥交换和数字签名（RSA）。组合系统（混合）使用非对称加密来共享对称会话密钥，然后使用对称加密进行数据传输。这出现在 SSL/TLS 中。理解这些角色可以避免错误。
12. Data Integrity and Verification: Parity vs Checksum vs ARQ | 数据完整性与验证：奇偶校验 vs 校验和 vs 自动重复请求
Students often confuse parity checks, checksums, and echo checks. A single parity bit can detect an odd number of bit errors, but cannot correct them. Many answer that a parity check can correct errors – it cannot. Checksums work by summing data blocks and sending the sum; the receiver recalculates and compares. If they match, data is assumed correct. However, checksums cannot identify which bit is wrong, so correction is not possible without retransmission. Automatic Repeat reQuest (ARQ) is a protocol that uses acknowledgements and timeouts to request retransmission when an error is detected by parity or checksum.
学生经常混淆奇偶校验、校验和与回波检验。单个奇偶位可以检测奇数个比特错误，但不能纠正它们。许多人回答奇偶校验可以纠正错误——它不能。校验和通过将数据块相加并发送总和来工作；接收方重新计算并比较。如果匹配，则假定数据正确。然而，校验和无法识别哪个比特出错，因此在没有重传的情况下无法纠正。自动重复请求 (ARQ) 是一种协议，使用确认和超时机制，当奇偶校验或校验和检测到错误时请求重传。
Exam pitfall: ‘Explain how a checksum can correct errors.’ This is incorrect; only error‑correcting codes like Hamming code can correct. The correct answer explains that checksum only detects, then ARQ requests retransmission. Another mistake: stating that even parity sets the parity bit so that the number of 1s is even; students sometimes write ‘makes the number of 0s even’. Careful with wording.
考试陷阱：“解释校验和如何纠正错误。”这是错误的；只有像汉明码这样的纠错码才能纠正。正确答案应解释校验和仅能检测错误，然后由 ARQ 请求重传。另一个错误：称偶校验将校验位设置为使得“1 的个数为偶数”；学生有时会写成“使 0 的个数为偶数”。注意措辞。
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