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一、测试优先
（1）为函数写一个规格说明。
（2）为上一步的规格说明写一些测试用例。
（3）编写实际代码。一旦你的代码通过了所有你写的测试用例，这个函数就算完成了。
二、分区测试（三个例子）
EX1：
/**
* @param val another BigInteger
* @return a BigInteger whose value is (this * val).
*/
public BigInteger multiply(BigInteger val)测试：a、b如何取值
BigInteger a = ...;
BigInteger b = ...;
BigInteger ab = a.multiply(b);方案：
一共7*7个区域
- EX2
/**
* @param a an argument
* @param b another argument
* @return the larger of a and b.
*/
public static int max(int a, int b)分析区域：
*a与b的关系 a < b a = b a > b
*a的值 a = 0 a < 0 a > 0 a = 最小的整数 a = 最大的整数
*b的值 b = 0 b < 0 b > 0 b = 最小的整数 b = 最大的整数
- EX3
/**
* Reverses the end of a string.
*
* 012345 012345
* For example: reverseEnd("Hello, world", 5) returns "Hellodlrow ,"
* <-----> <----->
*
* With start == 0, reverses the entire text.
* With start == text.length(), reverses nothing.
*
* @param text non-null String that will have its end reversed
* @param start the index at which the remainder of the input is reversed,
* requires 0 <= start <= text.length()
* @return input text with the substring from start to the end of the string reversed
*/
public static String reverseEnd(String text, int start)start测试分区：
A start = 0, start = 5, start = 100
B start < 0, start = 0, start > 0
C start = 0, 0 < start < text.length(), start = text.length()
D start < text.length(), start = text.length(), start > text.length()
注：选第三个，因为第四个不是输入的范围内
text测试分区
A text contains some letters; text contains no letters, but some numbers; text contains neither letters nor numbers
B text.length() = 0; text.length() > 0
C text.length() = 0; text.length()-start is odd; text.length()-start is even
D text is every possible string from length 0 to 100
注：选择B、C 。选择C是因为，例如(abc def ,2)
翻转后变成 abc fed(e的位置没有改变)，这可能需要特殊的行为来处理，也可能是bug产生的原因
三、如何写测试
/**
* Reverses the end of a string.
*
* For example:
* reverseEnd("Hello, world", 5)
* returns "Hellodlrow ,"
*
* With start == 0, reverses the entire text.
* With start == text.length(), reverses nothing.
*
* @param text non-null String that will have
* its end reversed
* @param start the index at which the
* remainder of the input is
* reversed, requires 0 <=
* start <= text.length()
* @return input text with the substring from
* start to the end of the string
* reversed
*/
static String reverseEnd(String text, int start)我们应该在测试时记录下我们的测试策略，例如我们是如何分区的，有哪些特殊值、边界值等等
/*
* Testing strategy
*
* Partition the inputs as follows:
* text.length(): 0, 1, > 1
* start: 0, 1, 1 < start < text.length(),
* text.length() - 1, text.length()
* text.length()-start: 0, 1, even > 1, odd > 1
*
* Include even- and odd-length reversals because
* only odd has a middle element that doesn't move.
*
* Exhaustive Cartesian coverage of partitions.
*/另外，每一个测试方法都要有一个小的注解，说明这个测试方法是代表测试策略中的哪一部分，例如：
// covers test.length() = 0,
// start = 0 = text.length(),
// text.length()-start = 0
@Test public void testEmpty() {
assertEquals("", reverseEnd("", 0));
}四、黑盒测试与白盒测试
黑：只依据函数的规格说明来选择测试用例，而不关心函数是如何实现
白：考虑实现的方法
例题：Q???
五、自动化测试、回归测试
自动化测试（Automated testing）：是指自动地运行测试对象，输入对应的测试用例，并记录结果的测试
回归测试：我们称修改代码带来新的bug的现象为“回归”，而在修改后重新运行所有的测试
- 练习
什么情况下应该重新运行所有的 JUnit 测试? - 在使用 git add/commit/push之前
- 在优化一个函数的性能后
- 在使用覆盖率工具时
- 在修改一个bug后
注：全部。第三个：Rerunning tests is an essential part of using a code coverage tool, because you want to see the code lines that your tests don’t reach.