AP CSA 1.6：复合赋值运算符
In the previous lessons, we used assignment statements like x = x + 1. In this lesson, we learn shorter ways to update a variable’s value.
前几课我们写过像 x = x + 1 这样的赋值语句。这一课，我们学习更简洁的变量更新写法。
Compound assignment operators update a variable by doing math and assignment in one step. 复合赋值运算符把“数学运算”和“重新赋值”合成一步完成。
Core Concepts:
核心知识点
1. Main Idea
核心理解
A compound assignment operator is a shortcut for updating a variable.
复合赋值运算符是“更新变量”的简写。
x += 5;
means:
x = x + 5;
The variable on the left is used in the calculation, and then the new result is stored back into the same variable.
左边的变量先参与计算，然后新的结果再存回这个变量。
2. Compound Assignment Operators
复合赋值运算符
| Operator | Meaning | Same As | Example Result |
|---|---|---|---|
+= |
add and assign | x = x + n |
x += 3 |
-= |
subtract and assign | x = x - n |
x -= 3 |
*= |
multiply and assign | x = x * n |
x *= 3 |
/= |
divide and assign | x = x / n |
x /= 3 |
%= |
remainder and assign | x = x % n |
x %= 3 |
Example:
int score = 10;
score += 5;
System.out.println(score);
score *= 2;
System.out.println(score);
Output:
15
30
score += 5 changes score from 10 to 15.
score *= 2 changes score from 15 to 30.
score += 5 把 score 从 10 改成 15。
score *= 2 把 score 从 15 改成 30。
3. Increment and Decrement
自增与自减
When you only want to add or subtract 1, Java has even shorter operators.
如果你只想加 1 或减 1，Java 有更短的写法。
| Code | Same As | 中文理解 |
|---|---|---|
x++ |
x = x + 1 |
x 增加 1 |
x-- |
x = x - 1 |
x 减少 1 |
Example:
int count = 0;
count++;
count++;
count--;
System.out.println(count);
Output:
1
Step by step:
| Line | count |
|---|---|
int count = 0; |
0 |
count++; |
1 |
count++; |
2 |
count--; |
1 |
4. AP Exam Focus: Postfix Only
AP 考点：只需掌握简单后缀形式
You may see x++ and x-- on the AP exam.
AP 考试中可能会看到 x++ 和 x--。
You do not need to worry about tricky prefix expressions like this:
++x
You also do not need to evaluate tricky expressions like this:
System.out.println(x++);
For AP CSA, focus on the simple meaning:
在 AP CSA 中，重点掌握简单含义：
x++;
means:
x = x + 1;
5. Code Tracing
代码追踪
Code tracing means simulating the program line by line, like you are the computer.
代码追踪就是像电脑一样，一行一行模拟代码执行。
Example:
int x = 0;
int y = 1;
int z = 2;
x--;
y++;
z += y;
Trace table:
| Line | x |
y |
z |
|---|---|---|---|
| Start | 0 | 1 | 2 |
x--; |
-1 | 1 | 2 |
y++; |
-1 | 2 | 2 |
z += y; |
-1 | 2 | 4 |
Final values:
x = -1, y = 2, z = 4
The key is to update the table immediately after each line.
关键是：每执行一行，就马上更新变量表。
6. Integer Division Reminder
整数除法提醒
Compound operators still follow normal Java rules.
复合赋值运算符仍然遵守 Java 的普通规则。
int y = 5;
y /= 2;
System.out.println(y);
Output:
2
Because y is an int, 5 / 2 gives 2, not 2.5.
因为 y 是 int，所以 5 / 2 的结果是 2，不是 2.5。
7. Common Beginner Mistakes
常见初学者错误
| Mistake | Wrong Code | Why Wrong | Fixed Code | 中文解释 |
|---|---|---|---|---|
Using =+ instead of += |
x =+ 5; |
This assigns positive 5 to x |
x += 5; |
=+ 不是“加后赋值” |
| Forgetting the variable changes | x += 3; then still thinking x is old value |
x has been updated |
Trace the new value | 执行后变量值已经改变 |
Treating x++ as adding 2 |
x++; |
It only adds 1 | x += 2; |
++ 只加 1 |
Expecting decimal from int division |
int x = 5; x /= 2; |
int / int keeps integer result |
Use double if needed |
整数除法会舍掉小数部分 |
8. Debugging Example
调试例子
Buggy code:
int score = 10;
score =+ 5;
System.out.println(score);
Output:
5
Problem: =+ is not the same as +=.
问题：=+ 和 += 不是一回事。
Fixed code:
int score = 10;
score += 5;
System.out.println(score);
Output:
15
| Bug | Type | Fix |
|---|---|---|
score =+ 5; |
Logic error | Use score += 5; |
score += ; |
Syntax error | Put a value after += |
score /= 0; |
Runtime error | Do not divide by zero |
A logic error may still run, but the answer is wrong. A syntax error prevents the program from compiling. A runtime error happens while the program is running.
逻辑错误可能能运行，但结果错。语法错误会导致程序无法编译。运行时错误是在程序运行过程中出错。
9. Mini Practice
小练习
Practice 1: Choose the Output
练习 1：选择输出
What is printed?
int x = 4;
x += 3;
x *= 2;
System.out.println(x);
Answer:
14
Explanation: 4 + 3 = 7, then 7 * 2 = 14.
解释：先 4 + 3 = 7，再 7 * 2 = 14。
Practice 2: Predict the Result
练习 2：预测结果
What are the final values of x, y, and z?
int x = 3;
int y = 5;
int z = 2;
x = z * 2;
y = y / 2;
z++;
Answer:
x = 4, y = 2, z = 3
Explanation: z * 2 is 4; 5 / 2 is 2 because this is integer division; z++ changes z from 2 to 3.
解释：z * 2 是 4；5 / 2 是 2，因为这是整数除法；z++ 把 z 从 2 改成 3。
Practice 3: Fix the Code
练习 3：修复代码
The goal is to add 5 to score. Fix the code.
int score = 10;
score =+ 5;
System.out.println(score);
Fixed code:
int score = 10;
score += 5;
System.out.println(score);
Output:
15
Explanation: use += when you want to add to the current value and store the result back.
解释：如果要在原来的值上加，并把结果存回变量，要用 +=。
Quick Checklist
快速检查清单
Before answering a compound assignment question, check:
做复合赋值题前，检查：
- What is the variable’s current value?
- Is the operator
+=,-=,*=,/=, or `%=? - Does the statement update the variable on the left?
- Is it
+=, not=+? - Does
x++mean add exactly1? - Does
x--mean subtract exactly1? - Are you tracing the code line by line?
- If division is used, is it integer division?
- Did you update the trace table after every line?