C# pass any method as a parameter

cmethodsparameters

When logging, you always get entangled in string literals.

I solved that nicely for properties, fields and variables by passing an Expression<Func<T>> expression (as explained here), so you can do things like this:

public void Demo(string someArgument)
{
    LogFrameWork.LogLine("Demo"); // goal is to get rid of these string literals
    LogFramework.Log(() => someArgument);
}

I want to do something similar for the method Demo itself:

public void Demo(string someArgument)
{
    LogFramework.Log(this.Demo);
}

I tried things like this:

public static void Log(Delegate method)
{
    string methodName = method.Method.Name;
    LogLine(methodName);
}

and this:

public static void Log(Action method)
{
    string methodName = method.Method.Name;
    LogLine(methodName);
}

But I get compiler errors like these:

Argument 1: cannot convert from 'method group' to 'System.Delegate' 
Argument 1: cannot convert from 'method group' to 'System.Action'

I could introduce a bunch of overloads using Func<…> and Action<…>, but that sounds overly complex.

Is there a way to cover this for any method with any number of parameters and an optional result?

–jeroen

PS: I think this question might have some relevance here, but no answers that got me a 'aha' feeling 🙂

Best Answer

You can also achieve this without using ExpressionTrees through System.Diagnostics.StackTrace.

StackTrace trace = new StackTrace();

And then:

trace.GetFrame(0).GetMethod().Name

To get the MethodInfo and then name of the current method, or:

trace.GetFrame(1).GetMethod().Name

To get the calling method.

Related Solutions

How to pass a function as a parameter in C

Declaration

A prototype for a function which takes a function parameter looks like the following:

void func ( void (*f)(int) );

This states that the parameter f will be a pointer to a function which has a void return type and which takes a single int parameter. The following function (print) is an example of a function which could be passed to func as a parameter because it is the proper type:

void print ( int x ) {
  printf("%d\n", x);
}

Function Call

When calling a function with a function parameter, the value passed must be a pointer to a function. Use the function's name (without parentheses) for this:

func(print);

would call func, passing the print function to it.

Function Body

As with any parameter, func can now use the parameter's name in the function body to access the value of the parameter. Let's say that func will apply the function it is passed to the numbers 0-4. Consider, first, what the loop would look like to call print directly:

for ( int ctr = 0 ; ctr < 5 ; ctr++ ) {
  print(ctr);
}

Since func's parameter declaration says that f is the name for a pointer to the desired function, we recall first that if f is a pointer then *f is the thing that f points to (i.e. the function print in this case). As a result, just replace every occurrence of print in the loop above with *f:

void func ( void (*f)(int) ) {
  for ( int ctr = 0 ; ctr < 5 ; ctr++ ) {
    (*f)(ctr);
  }
}

Source

Java – Is Java “pass-by-reference” or “pass-by-value”

Java is always pass-by-value. Unfortunately, when we deal with objects we are really dealing with object-handles called references which are passed-by-value as well. This terminology and semantics easily confuse many beginners.

It goes like this:

public static void main(String[] args) {
    Dog aDog = new Dog("Max");
    Dog oldDog = aDog;

    // we pass the object to foo
    foo(aDog);
    // aDog variable is still pointing to the "Max" dog when foo(...) returns
    aDog.getName().equals("Max"); // true
    aDog.getName().equals("Fifi"); // false
    aDog == oldDog; // true
}

public static void foo(Dog d) {
    d.getName().equals("Max"); // true
    // change d inside of foo() to point to a new Dog instance "Fifi"
    d = new Dog("Fifi");
    d.getName().equals("Fifi"); // true
}

In the example above aDog.getName() will still return "Max". The value aDog within main is not changed in the function foo with the Dog "Fifi" as the object reference is passed by value. If it were passed by reference, then the aDog.getName() in main would return "Fifi" after the call to foo.

Likewise:

public static void main(String[] args) {
    Dog aDog = new Dog("Max");
    Dog oldDog = aDog;

    foo(aDog);
    // when foo(...) returns, the name of the dog has been changed to "Fifi"
    aDog.getName().equals("Fifi"); // true
    // but it is still the same dog:
    aDog == oldDog; // true
}

public static void foo(Dog d) {
    d.getName().equals("Max"); // true
    // this changes the name of d to be "Fifi"
    d.setName("Fifi");
}

In the above example, Fifi is the dog's name after call to foo(aDog) because the object's name was set inside of foo(...). Any operations that foo performs on d are such that, for all practical purposes, they are performed on aDog, but it is not possible to change the value of the variable aDog itself.

For more information on pass by reference and pass by value, consult the following SO answer: https://stackoverflow.com/a/430958/6005228. This explains more thoroughly the semantics and history behind the two and also explains why Java and many other modern languages appear to do both in certain cases.

Best Answer

Related Solutions

How to pass a function as a parameter in C

Java – Is Java “pass-by-reference” or “pass-by-value”

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