CODEWITHSUNDEEP
c#.netlazy-loading
f0rt
f0rt

Reputation: 1921

How to debug Lazy<T>?

I have a Lazy<T> initialized with a lambda. How to see the body of the initializing lambda while debugging? I expect to have something like the DebugView of the Expression class but I've found nothing like that.

Upvotes: 4

Views: 1531

Answers (1)

MarcinJuraszek
MarcinJuraszek

Reputation: 125620

Because Lazy<T> takes a delegate, there is no Expression class involved. Your lambda is compiled like any other code in your project and there is no preview of that code during debug.

Lambda expression can be compiled either into IL or transformed into Expression Tree. Which one happens depends on the context. If your parameter is declared as delegate regular IL code will be generated. If it's Expression<TFunc> you'll get expression tree which can be previewed.

It's nicely explained on MSDN, based on Where method, which has two versions: Enumerable.Where which takes Func<T, bool> and Queryable.Where which takes Expression<Func<T, bool>>.

When you use method-based syntax to call the Where method in the Enumerable class (as you do in LINQ to Objects and LINQ to XML) the parameter is a delegate type System.Func<T, TResult>. A lambda expression is the most convenient way to create that delegate. When you call the same method in, for example, the System.Linq.Queryable class (as you do in LINQ to SQL) then the parameter type is an System.Linq.Expressions.Expression<Func> where Func is any of the Func delegates with up to sixteen input parameters. Again, a lambda expression is just a very concise way to construct that expression tree. The lambdas allow the Where calls to look similar although in fact the type of object created from the lambda is different.

Upvotes: 4

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