C# Passing Function as Argument [duplicate]

I've written a function in C# that does a numerical differentiation. It looks like this:

public double Diff(double x)
{ double h = 0.0000001; return (Function(x + h) - Function(x)) / h;
}

I would like to be able to pass in any function, as in:

public double Diff(double x, function f)
{ double h = 0.0000001; return (f(x + h) - f(x)) / h;
}

I think this is possible with delegates (maybe?) but I'm not sure how to use them.

3 Answers

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));

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);
}

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

Usage:

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