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Home/ Questions/Q 7174481
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Editorial Team
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Editorial Team
Asked: May 28, 20262026-05-28T16:03:48+00:00 2026-05-28T16:03:48+00:00

This is a bit of an odd one. I’m trying to stub a method

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This is a bit of an odd one. I’m trying to stub a method which has out parameters, I don’t care about what the parameters are so I’m ignoring the arguments. It looks like this:

List<Foo> ignored;
A.CallTo(() => fake.Method(out ignored))
  .Returns(something);

This works without any problems when the stubbed method is called like so:

List<Foo> target;
var result = service.Method(out target);

However, it doesn’t work when the target is pre-initialised. For example:

List<Foo> target = new List<Foo>();
var result = service.Method(out target);

When I inspect the Tag on the fake, I can see that the out parameters are being recorded as <NULL> so I suspect they’re not matching when the out target is already set to something. I’ve tried setting the ignored in my test to new List<Foo>() and also tried A<List<Foo>>.Ignored but neither has any effect on the result.

So my question is, does anyone know how to stub a method with out parameters if the out parameter target already has a value?

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  1. Editorial Team
    Editorial Team
    2026-05-28T16:03:50+00:00Added an answer on May 28, 2026 at 4:03 pm

    Update: since FakeItEasy 1.23.0 the initial value of out parameters is ignored when matching, so no need for WithAnyArguments

    , Five minutes later and I’ve found an acceptable solution (in this scenario). As I’m not interested in what arguments are passed to this method, so if I use the WithAnyArguments() method then it seems to work; this must shortcut the argument checking all together, I guess.

    The final code is:

    List<Foo> ignored;
    A.CallTo(() => fake.Method(out ignored))
      .WithAnyArguments()
      .Returns(something);
    

    This obviously doesn’t solve the problem if I don’t want to ignore all the arguments. I’ll only accept this answer if nobody has a more sophisticated solution.

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