Question

我想通过 FluentAssertion 语法检查方法的返回值。请考虑以下代码段：

public interface IFoo
{
    Task<int> DoSomething();
}

public class Bar
{
    private readonly IFoo _foo;
    private static int _someMagicNumber = 17;

    public Bar(IFoo foo)
    {
        _foo = foo;
    }

    public async Task<int> DoSomethingSmart()
    {
        try
        {
            return await _foo.DoSomething();
        }
        catch
        {
            return _someMagicNumber;
        }
    }
}

[TestFixture]
public class BarTests
{
    [Test]
    public async Task ShouldCatchException()
    {
        // Arrange
        var foo = Substitute.For<IFoo>();
        foo.DoSomething().Throws(new Exception());
        var bar = new Bar(foo);
        Func<Task> result = () => bar.DoSomethingSmart();

        // Act-Assert
        await result.Should().NotThrowAsync();
    }

    [Test]
    public async Task ShouldReturnDefaultValueWhenExceptionWasThrown()
    {
        // Arrange
        var foo = Substitute.For<IFoo>();
        foo.DoSomething().Throws(new Exception());
        var bar = new Bar(foo);

        // Act
        var result = await bar.DoSomethingSmart();

        // Assert
        result.Should().Be(17);
    }
}

我的目标是将这两个测试合并为新的测试，但我想保留流畅的断言检查：result.Should().NotThrowAsync();

所以我的问题是在示例中如何检查返回值是17的第一个测试？

Answer 1

当前版本的Fluent断言（5.5.3）不能区分Repetition A regular expression may be followed by one of several repetition operators: --------------------------------------------------------------- ? The preceding item is optional and matched at most once. --------------------------------------------------------------- * The preceding item will be matched zero or more times. + The preceding item will be matched one or more times. {n} The preceding item is matched exactly n times. {n,} The preceding item is matched n or more times. {,m} The preceding item is matched at most m times. This is a GNU extension. {n,m} The preceding item is matched at least n times, but not more than m times.和Func<Task>。两种类型都由Func<Task<T>>处理，AsyncFunctionAssertions将这两种类型分配给Func<Task>，因此失去了Task<T>的返回值。

一种避免这种情况的方法是将返回值分配给局部变量。

[Test]
public async Task ShouldCatchException()
{
    // Arrange
    var foo = Substitute.For<IFoo>();
    foo.DoSomething().Throws(new Exception());
    var bar = new Bar(foo);

    // Act
    int? result = null;
    Func<Task> act = async () => result = await bar.DoSomethingSmart();

    // Act-Assert
    await act.Should().NotThrowAsync();
    result.Should().Be(17);
}

我已经在Fluent Assertion问题跟踪器上创建了issue。

用FluentAssertion异常语法检查返回值

1 个答案: