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".
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.
Is there something I'm overlooking here?
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
}
And here's a way to do this with AutoFixture.Xunit:
[Theory, AutoData]
public void GenerateTwoDistinctNonZeroIntegersWithAutoFixtureXunit(
Generator<int> generator)
{
var numbers = generator.Where(x => x != 0).Distinct().Take(2);
// -> 72, 117
}
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