CODEWITHSUNDEEP
c#lambda
Tachyon
Tachyon

Reputation: 2411

Simple generic call with return type

I am currently trying to create a generic call that takes in a value that needs to be returned by the function that is being passed in. It would appear to be a simple generic method, but I for the life of me cannot figure out why it isn't working.

What I am looking to do is pass a method onto this method, so that it can call and API and handle any errors that it comes across. I want to pass a gRPC call across to the method, wrap it in a try catch then invoke and catch any errors. I figured a generic would be the best way to accomplish this but it doesn't seem to be working. If someone has a recommendation to what will work better, I am all ears.

The generic call I came up with is super simple:

public T MakeCall<T>(Func<T> calling)
{
    var res = calling.Invoke();


    return res;
}

The calling code is as follows

static void Main(string[] args)
{
    var caller = new Caller();

    caller.MakeCall<int>(Add(1, 2));

    Thread.Sleep(Timeout.Infinite);
}

public static int Add(int one, int two)
{
    return one + two;
}

It is giving me an error that says,

"int is not assignable to parameter type System.Func"

Upvotes: 0

Views: 65

Answers (2)

miechooy
miechooy

Reputation: 3422

Because Add() returns int whereas int is not Func<T>. You can use () => Add(1,2) as a parameter.

So the solution would be 

caller.MakeCall<int>(() => Add(1, 2));

Upvotes: 2

Dmitrii Bychenko
Dmitrii Bychenko

Reputation: 186668

Well,

MakeCall<int>(Add(1, 2))

means "execute Add(1, 2) and then call MakeCall<int> with the result (which is int)". You should put a Func<int>, not int:

MakeCall<int>(() => Add(1, 2))

Here

() => Add(1, 2)

is a Func<int> which takes nothing and returns Add(1, 2) result

Upvotes: 4

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