在异步委托中断言异常

Question

我使用的是 NUnit 3。我写了一个扩展方法：

public static T ShouldThrow<T>(this TestDelegate del) where T : Exception {
  return Assert.Throws(typeof(T), del) as T;
}

这允许我这样做：

TestDelegate del = () => foo.doSomething(null);
del.ShouldThrow<ArgumentNullException>();

现在我想要类似的异步：

AsyncTestDelegate del = async () => await foo.doSomething(null);
del.ShouldThrowAsync<ArgumentNullException>();

所以我写了这个：

public static async Task<T> ShouldThrowAsync<T>(this AsyncTestDelegate del) where T : Exception {
  return (await Assert.ThrowsAsync(typeof(T), del)) as T;
}

但这不起作用：'Exception' does not contain a definition for 'GetAwaiter' and no extension method 'GetAwaiter' accepting a first argument of type 'Exception' could be found (are you missing a using directive or an assembly reference?)。

我做错了什么？

Answer 1

据我所知，Assert.ThrowsAsync 不会返回 Task，因此无法等待。从您的扩展方法中删除 await。

public static T ShouldThrowAsync<T>(this AsyncTestDelegate del) where T : Exception {
  return Assert.ThrowsAsync(typeof(T), del) as T;
}

来自docs 的示例用法。请注意，Assert.ThrowsAsync 返回一个 MyException 并且 await 在委托中。

[TestFixture]
public class UsingReturnValue
{
  [Test]
  public async Task TestException()
  {
    MyException ex = Assert.ThrowsAsync<MyException>(async () => await MethodThatThrows());

    Assert.That( ex.Message, Is.EqualTo( "message" ) );
    Assert.That( ex.MyParam, Is.EqualTo( 42 ) ); 
  }
}