Setup multiple results based on given args, plus an additional case to fallback invocations

[amiralles]

Hi guys i have this code,

public class Demo {
        public interface IFoo {
            int Bar(string arg);
        }
 
        [Test]
        public void TestFoo() {
            var foo = new Mock<IFoo>();
            foo.Setup(f => f.Bar("aaa")).Returns(1);
            foo.Setup(f => f.Bar("bbb")).Returns(2);
            //
            foo.Setup(f => f.Bar(It.IsAny<string>())).Returns(3);
            //instead of the previous line, I would go with something like this:
            //foo.Setup(f => f.Bar(It.IsAnyOther<string>())).Returns(3);
 
            //With the current setup, these first two asserts are gonna fail,
            //because It.IsAny<string>() is going to match for any string argument....
            Assert.AreEqual(1, foo.Object.Bar("aaa"));
            Assert.AreEqual(2, foo.Object.Bar("bbb"));
            Assert.AreEqual(3, foo.Object.Bar("ccc"));
        }
    }

What I want, is to provide a default result for any invocation that does not match against the specified setup.
I know if I remove It.IsAny<string>  I'll get the default value of the method's return type, which in the sample above will be 0, but I want to be 3.
Is there any hack or known workaround to do that?
Thanks!

[Daniel Cazzulino]

Set the more generic setup first. The other two will override it afterwards.

/kzu

--
Daniel Cazzulino

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[mari...@gmail.com]

Change the order of the setups definition to:

 

  foo.Setup(f => f.Bar(It.IsAny<string>())).Returns(3);

  foo.Setup(f => f.Bar("aaa")).Returns(1);

  foo.Setup(f => f.Bar("bbb")).Returns(2);

 

If that does not work you can use It.Is matcher instead:

 

foo.Setup(f => f.Bar(It.Is(a => a != “aaa” && a != “bbb”)>())).Returns(3);

 

Regards

 

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[Ale Miralles]

Thanks, worked like  a charm!