Explainer · 8.2.2

讲解 · 8.2.2

Why use an array at all

为什么要使用数组

Isla wants to store one exam mark for each of 30 students. Without an array she needs 30 separate variables: Mark1, Mark2, all the way to Mark30. To add them up she writes 30 lines. To change the class size she rewrites the lot. It works, but nothing about it scales.

Isla 想为 30 名学生各存储一个考试成绩。如果不用数组,她需要 30 个独立的变量:Mark1、Mark2,一直到 Mark30。要把它们加起来,她得写 30 行代码。要改变班级人数,她得把这些全部重写。这样做行得通,但完全无法扩展。

An array replaces all 30 with one named structure. Every value sits in a numbered slot, and one loop can visit every slot in turn.

一个数组用一个命名的结构取代了全部 30 个变量。每一个值都存放在一个编号的槽位中,一个循环就能依次访问每一个槽位。

Explainer · 8.2.1

讲解 · 8.2.1

Declaring a 1D array

声明一维数组

A declaration tells the program the array's name, how many elements it holds, and what type of value goes in each one.

声明语句告诉程序数组的名称、它包含多少个元素,以及每个元素存放什么类型的值。

To reach one element you put an expression in the brackets. A fixed number reads one slot: StudentMark[1]. A variable as the index is what makes a loop work: StudentMark[Count] reads whichever slot Count currently points at, so a single line inside a loop can touch all 30.

要访问某一个元素,需要在方括号中放入一个表达式。固定的数字读取单个槽位:StudentMark[1]。而用变量作为索引才是循环得以运作的关键:StudentMark[Count] 读取的是 Count 当前所指向的槽位,因此循环内的一行代码就能访问全部 30 个元素。

Explainer · 8.2.1 + 8.2.3

讲解 · 8.2.1 + 8.2.3

2D arrays and nested iteration

二维数组与嵌套迭代

Think of a class register: one row of numbered seats, one value per seat. Now think of a spreadsheet: rows AND columns, and every cell needs two coordinates to find it, not one. That jump, from one number to two, is the entire difference between a 1D array and a 2D array.

想象一份班级名册:一排编号的座位,每个座位一个值。再想象一份电子表格:既有行又有列,每个单元格需要两个坐标才能定位,而不是一个。从一个数字到两个数字的这一跳跃,正是一维数组与二维数组之间的全部区别。

A 2D array is a grid, and every box needs two indices to find it: one for the row, one for the column.

一个二维数组是一个网格,每个格子都需要两个索引才能定位:一个是行,一个是列。

Nested iteration is how you visit every box in a 2D array: an outer loop walks the rows, an inner loop walks the columns of the current row. The inner loop finishes completely before the outer loop moves to the next row.

嵌套迭代是访问二维数组每个格子的方式:外层循环遍历行,内层循环遍历当前行的列。内层循环必须完全结束,外层循环才会移动到下一行。

Paper 2 scenarios lean on this shape every year: a grid of readings, one row per thing, one column per time slot (screen time per student per day, temperatures per station per week, hours per parking space per day). Here it is 50 spaces, 5 days, hours recorded in WeeklyUsage[Space, Day]. To total one space, the inner loop must finish all 5 days before the outer loop moves on.

试卷二的情境题年年都依赖这种结构:一个数据网格,每行对应一个对象,每列对应一个时间段(每个学生每天的屏幕使用时长、每个站点每周的气温、每个车位每天的时长)。这里是 50 个车位、5 天,时长记录在 WeeklyUsage[Space, Day] 中。要统计一个车位的总时长,内层循环必须先完成全部 5 天,外层循环才能继续。

// Total hours parked for each of the 50 spaces, across all 5 days
DECLARE Space : INTEGER
DECLARE Day : INTEGER
DECLARE TotalHours : INTEGER

FOR Space ← 1 TO 50
    TotalHours ← 0          // reset BEFORE the inner loop starts
    FOR Day ← 1 TO 5
        TotalHours ← TotalHours + WeeklyUsage[Space, Day]
    NEXT Day
    OUTPUT "Space ", Space, " total hours: ", TotalHours
NEXT Space

# Total hours parked for each of the 50 spaces, across all 5 days
for space in range(1, 51):
    total_hours = 0
    for day in range(1, 6):
        total_hours += WeeklyUsage[space][day]
    print("Space", space, "total hours:", total_hours)

// Total hours parked for each of the 50 spaces, across all 5 days
for (int space = 1; space <= 50; space++) {
    int totalHours = 0;
    for (int day = 1; day <= 5; day++) {
        totalHours += WeeklyUsage[space][day];
    }
    System.out.println("Space " + space + " total hours: " + totalHours);
}

' Total hours parked for each of the 50 spaces, across all 5 days
For space = 1 To 50
    totalHours = 0
    For day = 1 To 5
        totalHours = totalHours + WeeklyUsage(space, day)
    Next day
    Console.WriteLine("Space " & space & " total hours: " & totalHours)
Next space

Pseudocode or program code accepted here: the 15-mark scenario question is the one place in Paper 2 where Cambridge allows Python, Java or Visual Basic instead of pseudocode. Every other Paper 2 question (validation, trace tables, error-correction) stays pseudocode-only.

此处接受伪代码或程序代码:这道 15 分的情境题是试卷二中唯一允许用 Python、Java 或 Visual Basic 替代伪代码作答的地方。试卷二的其他题目(验证、追踪表、纠错)一律只接受伪代码。

Note the order inside the brackets never changes: it’s always [Space, Day] (or [space][day] in Python/Java), matching how WeeklyUsage was declared. Swap them and you read the wrong cell.

注意方括号内的顺序始终不变:总是 [Space, Day](在 Python/Java 中是 [space][day]),与 WeeklyUsage 的声明方式一致。顺序颠倒就会读到错误的单元格。

Explainer

讲解

How to trace code that indexes a 2D list

如何追踪索引二维列表的代码

Before answering a “what does this print?” question, trace it exactly like a trace table: one row per loop pass, tracking the index and the running total. Here it is worked in full on a fresh example: rainfall in mm, recorded at 3 stations over 3 months.

在回答"这段代码会打印什么?"这类问题之前,要像追踪表一样精确地追踪它:每次循环一行,记录索引和累计总数。这里以一个新例子完整演示:3 个站点、3 个月的降雨量(毫米)记录。

# RainfallMM[0] = Station A, [1] = Station B, [2] = Station C
RainfallMM = [
    [12, 8, 15],
    [20, 5, 9],
    [3, 11, 7]
]

total = 0
for month in range(3):
    total = total + RainfallMM[0][month]
print(total)

The row index (0, fixed) never changes. Only month moves, from 0 to 2. Output: 35.

行索引(0,固定)始终不变。只有 month 在变化,从 0 到 2。输出:35。