如何让AutoFixture创建一个大于0的整数，而不是另一个数字? [英] How to get AutoFixture create an integer that is >0, and not another number?

问题描述

我希望AutoFixture生成两个整数，而对于第二个整数，我不希望它为0或先前生成的数字.有没有办法告诉AutoFixture兑现要求".

I want AutoFixture to generate two integers, and for the second one, I don't want it to be 0, or the previous generated number. Is there a way to tell AutoFixture to honor that "requirement".

看着RandomNumericSequenceGenerator，我看起来像下限是1 ，因此我可能不必指定第一个要求.接下来，我正在查看播种"选项，但是如此答案所示，它将不会用于默认情况下是一个数字.

Looking at RandomNumericSequenceGenerator, I looks like the lower limit is 1, so I might not have to specify the first requirement. Next, I was looking at the "seeding" option, but as indicated in this answer, it won't be used for a number, by default.

这里有什么我可以俯瞰的地方吗?

Is there something I'm overlooking here?

推荐答案

这是使用普通AutoFixture做到这一点的一种方法:

Here's a way to do this with plain AutoFixture:

[Fact]
public void GenerateTwoDistinctNonZeroIntegersWithAutoFixture()
{
    var fixture = new Fixture();
    var generator = fixture.Create<Generator<int>>();

    var numbers = generator.Where(x => x != 0).Distinct().Take(2);
    // -> 72, 117
}

这是使用 AutoFixture.Xunit 的一种方法:

[Theory, AutoData]
public void GenerateTwoDistinctNonZeroIntegersWithAutoFixtureXunit(
    Generator<int> generator)
{
    var numbers = generator.Where(x => x != 0).Distinct().Take(2);
    // -> 72, 117
}

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