C# error when using explicitly defined Predicate

Question

i have a number of lambda expressions that will be using the same predicate in a where clause. As such i am using the predicate type for the first time. Here is what i have..

Predicate<Type> datePredicate = o => o.Date > DateTime.Now.AddDays(-1);

When i use it in my query (below), i am getting the following error..

Error:

The type arguments for method 'System.Linq.Enumerable.Where<TSource>(System.Collections.Generic.IEnumerable<TSource>, System.Func<TSource,int,bool>)' cannot be inferred from the usage. Try specifying the type arguments explicitly.

Usage:

Type t = collection.Where(datePredicate).SingleOrDefault();

Does anyone know what i am doing wrong?

How is your collection defined? – Josh M. May 10 '11 at 02:01 — Josh M., May 10 '11 at 02:01

score 3 · Accepted Answer · answered May 10 '11 at 02:08

3

Try this:

Func<MyObject, bool> datePredicate = (o => o.Date > DateTime.Now.AddDays(-1));

collection.Where(datepredicate);

Also when you're doing .SingleOrDefault(), not sure how that will magically turn into a Type since your List<T> is not a List<Type> as far as I know (since Type doesn't have a Date property).

answered May 10 '11 at 02:08

Josh M.

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thanks Josh your answer worked. can you tell me why i needed to use a Func and not a Predicate? – Grant May 10 '11 at 02:10
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This S/O question might provide some enlightenment: http://stackoverflow.com/questions/4317479/func-vs-action-vs-predicate – Russ Clarke May 10 '11 at 02:13
Just because `.Where` expects a `Func`. – Josh M. May 10 '11 at 02:15

score 0 · Answer 2 · answered May 10 '11 at 02:03

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The compiler cannot statically figure out what Type your o parameter is. I would presume o is of type DateTime. Compilers don't make presumptions :)

Predicate<DateTime> datePredicate = o => o.Date > DateTime.Now.AddDays(-1);

Try that.

answered May 10 '11 at 02:03

x0n

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I fixed my answer up - I think you edited your question while I was answering first. – x0n May 10 '11 at 02:06

C# error when using explicitly defined Predicate

2 Answers2