EasyMock：我如何在没有警告的情况下创建泛型类的模拟？ [英] EasyMock: How do I create a mock of a genericized class without a warning?

问题描述

代码

  private SomeClass< Integer> SomeClass的; 
 someClass = EasyMock.createMock（SomeClass.class）; 
  

给我一​​个警告类型安全性：SomeClass类型的表达式需要未经检查的转换符合SomeClass<整数>。

解决方案

AFAIK，当涉及类名字面值时，你无法避免未经检查的警告， code> SuppressWarnings 注解是处理这种情况的唯一方法。

请注意，缩小 SuppressWarnings 注释尽可能多。您可以将此批注应用于单个局部变量赋值：

  public void testSomething（）{
 
 @SuppressWarnings（unchecked）
 Foo<整数> foo = EasyMock.createMock（Foo.class）; 
 
 //其余的测试方法可能还会暴露其他警告
 
 code $ $ $ $ $ $ $ $ $ $ $ $ $ 使用辅助方法： 
 
 
  @SuppressWarnings（unchecked）
 private static< T> FOO< T> createFooMock（）{
 return（Foo< T>）EasyMock.createMock（Foo.class）; 
} 
 
 public void testSomething（）{
 Foo< String> foo = createFooMock（）; 
 
 //其余的测试方法仍可能会暴露其他警告
 
 code $ <$ pre 
The code
private SomeClass<Integer> someClass;
someClass = EasyMock.createMock(SomeClass.class);
gives me a warning "Type safety: The expression of type SomeClass needs unchecked conversion to conform to SomeClass<Integer>".
解决方案
AFAIK, you can't avoid the unchecked warning when a class name literal is involved, and the SuppressWarnings annotation is the only way to handle this.

Note that it is good form to narrow the scope of the SuppressWarnings annotation as much as possible. You can apply this annotation to a single local variable assignment:
public void testSomething() {

    @SuppressWarnings("unchecked")
    Foo<Integer> foo = EasyMock.createMock(Foo.class);

    // Rest of test method may still expose other warnings
}
or use a helper method:
@SuppressWarnings("unchecked")
private static <T> Foo<T> createFooMock() {
    return (Foo<T>)EasyMock.createMock(Foo.class);
}

public void testSomething() {
    Foo<String> foo = createFooMock();

    // Rest of test method may still expose other warnings
}


                        
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