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Home/ Questions/Q 152149

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Asked: May 11, 20262026-05-11T09:34:18+00:00 2026-05-11T09:34:18+00:00

I have used C# expressions before based on lamdas, but I have no experience

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I have used C# expressions before based on lamdas, but I have no experience composing them by hand. Given an Expression<Func<SomeType, bool>> originalPredicate, I want to create an Expression<Func<OtherType, bool>> translatedPredicate.

In this case SomeType and OtherType have the same fields, but they are not related (no inheritance and not based on a common interface).

Background: I have a repository implementation based on LINQ to SQL. I project the LINQ to SQL entities to my Model entities, to keep my model in POCO. I want to pass expressions to the repository (as a form of specifications) but they should be based on the model entities. But I can’t pass those expressions to the data context, since it expects expressions based on the LINQ to SQL entities.

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  1. 2026-05-11T09:34:19+00:00Added an answer on May 11, 2026 at 9:34 am

    With Expression, the simplest way is with a conversion expression:

    class Foo {     public int Value { get; set; } } class Bar {     public int Value { get; set; } } static class Program {     static void Main() {         Expression<Func<Foo, bool>> predicate =             x => x.Value % 2 == 0;         Expression<Func<Bar, Foo>> convert =             bar => new Foo { Value = bar.Value };          var param = Expression.Parameter(typeof(Bar), 'bar');         var body = Expression.Invoke(predicate,               Expression.Invoke(convert, param));         var lambda = Expression.Lambda<Func<Bar, bool>>(body, param);          // test with LINQ-to-Objects for simplicity         var func = lambda.Compile();         bool withOdd = func(new Bar { Value = 7 }),              withEven = func(new Bar { Value = 12 });     } } 

    Note however that this will be supported differently by different providers. EF might not like it, for example, even if LINQ-to-SQL does.

    The other option is to rebuild the expression tree completely, using reflection to find the corresponding members. Much more complex.

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