Setting property value on child instance to a fixed value with Autofixture

Question

Is it possible to assign a fixed value to a property on a child instance when building a parent with Autofixture? It will add default values to all the properties on the child instance like a charm, but I would like to override and assign a specific value to one of the properties on the child instance.

Given this parent/child relationship:

public class Person
{
    public string FirstName { get; set; }
    public string LastName { get; set; }

    public Address Address { get; set; }
}

public class Address
{
    public string Street { get; set; }
    public int Number { get; set; }
    public string City { get; set; }
    public string PostalCode { get; set; }
}

I would like to assign a specific value to the City property on the address instance. I was thinking in the lines of this test code:

var fixture = new Fixture();

var expectedCity = "foo";

var person = fixture
    .Build<Person>()
    .With(x => x.Address.City, expectedCity)
    .Create();

Assert.AreEqual(expectedCity, person.Address.City);

That is not possible. I guess, by the reflection exception

System.Reflection.TargetException : Object does not match target type.

...that Autofixture tries to assign the value to a City property on the Person instance instead of an Address instance.

Any suggestions?

And yes, I know that I could just add an extra step like the following:

var fixture = new Fixture();

var expectedCity = "foo";

// extra step begin
var address = fixture
    .Build<Address>()
    .With(x => x.City, expectedCity)
    .Create();
// extra step end

var person = fixture
    .Build<Person>()
    .With(x => x.Address, address)
    .Create();

Assert.AreEqual(expectedCity, person.Address.City);

...but was hoping for the first version or something similar (fewer steps, more terse).

Note: I'm using Autofixture v3.22.0

Answer 1

For completeness, here's another way to do it:

fixture.Customize<Address>(c => 
    c.With(addr => addr.City, "foo"));

var person = fixture.Create<Person>();

This will customize the creation of all instances of Address

If you end up using this often enough, it may be worthwhile wrapping it inside an ICustomization :

public class AddressConventions : ICustomization
{
    public void Customize(IFixture fixture)
    {
        fixture.Customize<Address>(c => 
            c.With(addr => addr.City, "foo"));
    }
}

fixture.Customize(new AddressConventions());

Answer 2

Not to be dismissive of the question, but the simplest solution might actually be this:

[Fact]
public void SimplestThingThatCouldPossiblyWork()
{
    var fixture = new Fixture();
    var expectedCity = "foo";
    var person = fixture.Create<Person>();
    person.Address.City = expectedCity;
    Assert.Equal(expectedCity, person.Address.City);
}

Assignment of explicit values to properties is something most languages already excel at (C# certainly does), so I don't think AutoFixture needs a complicated DSL to reproduce half of that functionality.

Answer 3

You can use a Do method on the builder:

var person = this.fixture
               .Build<Person>()
               .Do( x => x.Address.City = expectedCity )
               .Create();

You can also inject and freeze your instance to the fixture:

this.fixture.Inject( expectedCity );