C# Passing Function as Argument [duplicate]

Answer 1

Using the Func as mentioned above works but there are also delegates that do the same task and also define intent within the naming:

public delegate double MyFunction(double x);

public double Diff(double x, MyFunction f)
{
    double h = 0.0000001;

    return (f(x + h) - f(x)) / h;
}

public double MyFunctionMethod(double x)
{
    // Can add more complicated logic here
    return x + 10;
}

public void Client()
{
    double result = Diff(1.234, x => x * 456.1234);
    double secondResult = Diff(2.345, MyFunctionMethod);
}

Answer 2

public static T Runner<T>(Func<T> funcToRun)
{
    //Do stuff before running function as normal
    return funcToRun();
}

Usage:

var ReturnValue = Runner(() => GetUser(99));

Answer 3

There are a couple generic types in .Net (v2 and later) that make passing functions around as delegates very easy.

For functions with return types, there is Func<> and for functions without return types there is Action<>.

Both Func and Action can be declared to take from 0 to 4 parameters. For example, Func < double, int > takes one double as a parameter and returns an int. Action < double, double, double > takes three doubles as parameters and returns nothing (void).

So you can declare your Diff function to take a Func:

public double Diff(double x, Func<double, double> f) {
    double h = 0.0000001;

    return (f(x + h) - f(x)) / h;
}

And then you call it as so, simply giving it the name of the function that fits the signature of your Func or Action:

double result = Diff(myValue, Function);

You can even write the function in-line with lambda syntax:

double result = Diff(myValue, d => Math.Sqrt(d * 3.14));