.NET 4.0: Parameter naming inconsistency between Expression and MethodInfo

Question

Hello there,

I have just found some misterious behaviour while working with System.Linq.Expressions.Expression and System.Reflection.MethodInfo.

The code is as follows:

    static void Main(string[] args)
    {

        Expression<Func<double, double, double>> example = (x, y) => Math.Sin(x);

        //Prints out "x, y":
        Console.WriteLine(example.Parameters[0].Name + ", " + example.Parameters[1].Name);

        //Prints out "a":
        Console.WriteLine((example.Body as MethodCallExpression).Method.GetParameters()[0].Name);

    }

"a"? Where did my "x" go and where did this "a" come from?

Thinking that perhaps this is an alias used at low level, I have searched for "UsedName", "VisibleName" or something along those lines, but I haven't found anything.

Unfortunately, Expression does not feature a Parameters property (I believe only LambdaExpression does) which would return the "parameters in use", if any, in a given expression.

One can create a method that traverses the entire expression and collects the different parameters in use, but I was wondering if there is an easier way to do this.

Thanks a lot in advance.

Visual C# Express: 10.0.30319.1 RTMRel

.NET Framework: 4.0.30319 RTMRel

Answer 1

The a comes from the method you're calling - Math.Sin. x and y are parameters to your lambda expression; a is the parameter in Math.Sin.

If you want to get back x and y, cast example to LambdaExpression:

foreach (ParameterExpression p in example.Parameters)
{
    Console.WriteLine(p.Name); // Prints x then y
}

So no, there's no inconsistency here.

EDIT: If you want to find the arguments used in the method call - and you're sure it is just a method call, and the arguments will just be parameters - you can use something like:

var methodCall = (MethodCallExpression) example.Body;

// Implicitly casts each argument to ParameterExpression
foreach (ParameterExpression p in methodCall.Arguments)
{
    Console.WriteLine(p.Name);
}

In this case, it just prints out x as that's what's being used as the argument to Math.Sin.

Answer 2

Place the cursor over Sin in Visual Studio. Press the F12 key. Here's where a comes from. It comes from a developer in Redmond who decided to name it like this :-)

Answer 3

Thanks guys, that solves it (sorry, but I can't add comments or do anything after I closed my browser).