Output IEnumerable <T> and Take (x)?

Question

(using Entity Framework)

when I write:

IEnumerable<string> q = customers /*EF entity*/ .Select (c => c.Name.ToUpper()) .OrderBy (n => n) 

The C # compiler knows how to emit expression trees, which in turn are executed by sql:

 SELECT UPPER (Name) FROM Customer ORDER BY UPPER (Name) 

note also that there is an order by clause

but

I saw this link :

He wrote:

 IEnumerable<employee> emp = dc.Employees.Where(x => x.Desc.StartsWith("soft")); emp = emp.Take(1); 

after examining the final request that he saw:

 SELECT [t0].[Id], [t0].[Name], [t0].[Address], [t0].[Desc] AS [Desc] FROM [dbo].[Employee] AS [t0] WHERE [t0].[Desc] LIKE @p0 

Please note that there is no top offer

why?

shouldn't add Take(x) to the request?

will write it as follows:

 IEnumerable<employee> emp = (dc.Employees.Where(x => x.Desc.StartsWith("soft"))).Take(1); 

Would add a TOP clause to the query sent by SQL?

what's going on here?

(I already know that take not deferred execution)

Answer 1

Extension methods, such as Take() , are static methods, and therefore they are allowed at compile time.

The compile-time type emp is IEnumerable<employee> (because it is explicitly declared as one), so the compiler chooses Enumerable.Take instead of Queryable.Take , which does not perform any query translation.

If you were lazy and just used var instead of a type name:

 var emp = dc.Employees.Where(x => x.Desc.StartsWith("soft")); emp = emp.Take(1); 

this would work because the compiler would IQueryable<employee> for emp (because the expression you are initializing with this type), and thus Queryable.Take for the second call.

Answer 2

If you added Take(1) to the first expression, it will be on IQueryable<T> and thus added to SQL. But since you converted IQueryable<T> to IEnumerable<T> , Take(1) is executed in memory: IEnumerable<T> has an extension method Take() with the same name and, as far as the compiler knows, emp is IEnumerable<T> , and not IQueryable<T> , so the version in memory is called.