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Method called without passing parameters

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c#

I'm a beginner in C#. I encountered below code snippet in my project. I do not understand how ViewHelper.IsInView has been called without passing any parameters. Could anyone explain me this. Thanks in advance.

public static class ViewHelper
{
  public static bool IsInView(IFrameworkElement element)
  {
  ----------
  }
}

var Result = Views.Any(ViewHelper.IsInView);
like image 465
SKN Avatar asked Mar 09 '26 06:03

SKN


2 Answers

The Any method accepts a delegate - a pointer to a function - of the form Func<T, bool>. Meaning it expects a method that accepts an element of the type of the collection (I'm guessing IFrameworkElement in your case) and returns a bool - which is exactly the signature of the IsInView method.

The Any method then executes this delegate on elements in the Views collection until it encounters one that returns true.

In C#, there is an implicit conversion from a "method group" to a delegate type. Essentially, when you write

Views.Any(ViewHelper.IsInView)

It translates into

Views.Any(new Func<IFrameworkElement, bool>(ViewHelper.IsInView))
like image 174
Eli Arbel Avatar answered Mar 11 '26 19:03

Eli Arbel


What is being passed to Enumerable.Any is a delegate, the method is not being called at this point. If there are any views then Any will call that delegate with one or more of the views as the argument.

The delegate you're passing to Any has been created through something known as implicit method group conversion.

Views.Any is expecting a delegate of the type Func<IFrameworkElement, bool>, meaning it takes a single parameter of type IFrameworkElement and returns bool. You can create such a delegate from your method, as the signatures are compatible. This is how you would explicitly do this:

Func<IFrameworkElement, bool> predicate = 
    new Func<IFrameworkElement, bool>(ViewHelper.IsInView);

However, from C# 2.0 such a conversion can be done implicitly, meaning this code is exactly the same:

Func<IFrameworkElement, bool> predicate = ViewHelper.IsInView;
like image 38
Charles Mager Avatar answered Mar 11 '26 19:03

Charles Mager



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