这不会使用JUnit assertThat 方法签名：
public static< T> void assertThat（T actual，Matcher< T> matcher）编译器错误信息是：
错误：错误：行（102）找不到符号方法 assertThat（java.util.Map< java.lang.String ，java.lang.Class< java.util.Date>> ;, org.hamcrest.Matcher< java.util.Map< java.lang.String，java.lang.Class <？如果我改变 assertThat 方法签名为： public static< T> void assertThat（T result，Matcher< ;? extends T> matcher）然后编译工作。 p>
所以有三个问题：
作为参考，这里是的JUnit实现assertThat public static< T> void assertThat（T actual，Matcher< T> matcher）{ assertThat（，actual，matcher）; } public static< T> void assertThat（String reason，T actual，Matcher< T> matcher）{ if（！matcher.matches（actual））{描述description = new StringDescription（）; description.appendText（reason）; description.appendText（\\\Expected：）; matcher.describeTo（description）; 描述 .appendText（\\\ got：） .appendValue（actual） .appendText（\\\）; 抛出新的java.lang.AssertionError（description.toString（））; } }
解决方案
- 我必须引导您访问 www.angelikalanger/GenericsFAQ/JavaGenericsFAQ.html - 她做了一件了不起的工作。
基本想法是，您使用
< T extends SomeClass>当实际参数可以是 SomeClass 或
在你的例子中，
Map< / p>字符串，类<？扩展Serializable>>预期=空; Map< String，Class< java.util.Date>> result = null; assertThat（result，is（expected））;您的意思是 expected 可以包含表示实现 Serializable 的任何类的Class对象。你的结果图表示它只能容纳 Date 类对象。
当你传入结果时， T 到 Map 字符串到 Date 类对象，它与 Map 的 String 不匹配可序列化。
要检查的一件事 - 您确定要使用 Class< Date> ; 而不是日期？一般来说， String 到 Class
Given the following example (using JUnit with Hamcrest matchers):
Map<String, Class<? extends Serializable>> expected = null; Map<String, Class<java.util.Date>> result = null; assertThat(result, is(expected));This does not compile with the JUnit assertThat method signature of:
public static <T> void assertThat(T actual, Matcher<T> matcher)The compiler error message is:
Error:Error:line (102)cannot find symbol method assertThat(java.util.Map<java.lang.String,java.lang.Class<java.util.Date>>, org.hamcrest.Matcher<java.util.Map<java.lang.String,java.lang.Class <? extends java.io.Serializable>>>)However, if I change the assertThat method signature to:
public static <T> void assertThat(T result, Matcher<? extends T> matcher)Then the compilation works.
So three questions:
For reference, here is the JUnit implementation of assertThat:
public static <T> void assertThat(T actual, Matcher<T> matcher) { assertThat("", actual, matcher); } public static <T> void assertThat(String reason, T actual, Matcher<T> matcher) { if (!matcher.matches(actual)) { Description description = new StringDescription(); description.appendText(reason); description.appendText("\nExpected: "); matcher.describeTo(description); description .appendText("\n got: ") .appendValue(actual) .appendText("\n"); throw new java.lang.AssertionError(description.toString()); } }解决方案
First - I have to direct you to www.angelikalanger/GenericsFAQ/JavaGenericsFAQ.html -- she does an amazing job.
The basic idea is that you use
<T extends SomeClass>when the actual parameter can be SomeClass or any subtype of it.
In your example,
Map<String, Class<? extends Serializable>> expected = null; Map<String, Class<java.util.Date>> result = null; assertThat(result, is(expected));You're saying that expected can contain Class objects that represent any class that implements Serializable. Your result map says it can only hold Date class objects.
When you pass in result, you're setting T to exactly Map of String to Date class objects, which doesn't match Map of String to anything that's Serializable.
One thing to check -- are you sure you want Class<Date> and not Date? A map of String to Class<Date> doesn't sound terribly useful in general (all it can hold is Date.class as values rather than instances of Date)
As for genericizing assertThat, the idea is that the method can ensure that a Matcher that fits the result type is passed in.
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