C# - How can I pass a reference to a function that requires an out variable?

Question

public class Foo
{
    public void DoFoo()
    {
       int x;
       var coll = TheFunc("bar", out x);
    }

    public Func<string, int, ICollection<string>> TheFunc { get; set; }
}

Error: "Argument 2 should not be passed with the 'out' keyword."

public class Foo
{
    public void DoFoo()
    {
       int x;
       var coll = TheFunc("bar", out x);
    }

    public Func<string, out int, ICollection<string>> TheFunc { get; set; }
}

Error: "Invalid variance modifier. Only interface and delegate type parameters can be specified as variant."

How do I get an out parameter in this function?

see http://stackoverflow.com/questions/1365689/cannot-use-ref-or-out-parameter-in-lambda-expressions for some details — shelleybutterfly, Jul 22 '11 at 20:32
FYI the more general rule here is that *a type argument must be a type which is convertible to object*. "out int" is not convertible to object; you cannot convert "reference to an int variable" to object. — Eric Lippert, Jul 22 '11 at 20:35

score 8 · Answer 1 · answered Jul 22 '11 at 20:21

8

Define a delegate type:

public delegate ICollection<string> FooDelegate(string a, out int b);

public class Foo
{
    public void DoFoo()
    {
       int x;
       var coll = TheFunc("bar", out x);
    }

    public FooDelegate TheFunc { get; set; }
}

answered Jul 22 '11 at 20:21

cdhowie

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score 7 · Accepted Answer · answered Jul 22 '11 at 20:20

7

You need to make your own delegate:

delegate ICollection<string> MyFunc(string x, out int y);

answered Jul 22 '11 at 20:20

SLaks

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C# - How can I pass a reference to a function that requires an out variable?

2 Answers2