Is it possible to write a wrapper for the C# nameof operator?

Question

I read a related question and answer about the nameof operator, but it didn't help me, so I asked it here.

I want to write a wrapper for the C# nameof operator so not only will it will return the name of a class property, but also concatenate it with the class name.

Let's assume a class with single property:

class Foo
{
   public string SomeProperty {get; set;}
}

Now if compiling Console.WriteLine(nameof(Foo.SomeProperty)) with (C# 6 or higher), the result will be:

SomeProperty

So that is it possible to have something like this:

public string PrintFullName(???? object)
{
    //????
}

I put ???? for the input Type, because I don't know what the proper input Type is.

I want the result of the PrintFullName to be:

Foo.SomeProperty

I don't necessarily look for run-time solutions. Any compile-time workaround will also help.

You can't access Foo.SomeProperty since SomeProperty is an instance member of Foo. You can olny write instanceOfFoo.SomeProperty. You only may use reflexion on the class type to get members definitions. — user12031933
– user12031933, Commented Sep 25, 2019 at 15:28
@OlivierRogier Console.WriteLine(nameof(Foo.SomeProperty)) is now a valid C# expression for C# 6 — yekanchi
– yekanchi, Commented Sep 25, 2019 at 15:30
It's still not clear - assuming that you got this to work, why would PrintFullName return "Foo.SomeProperty"? What is the connection between the two? — Scott Hannen
– Scott Hannen, Commented Sep 25, 2019 at 15:30
Maybe you meant something like Console.WriteLine(typeof(Foo).Name + "." + nameof(Foo.SomeProperty)); ? — Matthew Watson
– Matthew Watson, Commented Sep 25, 2019 at 15:31
@MatthewWatson it prints the result but is it possible to wrap it inside something that inputs the property alone? — yekanchi
– yekanchi, Commented Sep 25, 2019 at 15:32

Peter Mortensen · Accepted Answer · 2019-09-25 22:51:11Z

2

Sure this is possible, using expression trees.

A full explanation on this site (all credits to Dave Glick).

It comes down to this:

public void UseNames(string className, string memberName)
{
    // Your code to use the class & membername go here
}

public void UseNames<T>(Expression<Func<T, object>> expression)
{
    MemberExpression member = expression.Body as MemberExpression;
    if (member == null)
    {
        // The property access might be getting converted to object to match the func
        // If so, get the operand and see if that's a member expression
        member = (expression.Body as UnaryExpression)?.Operand as MemberExpression;
    }
    if (member == null)
    {
        throw new ArgumentException("Action must be a member expression.");
    }

    // Pass the names on to the string-based UseNames method
    UseNames(typeof(T).Name, member.Member.Name);
}

public void UseNames<T>(Expression<Func<T, string>> expression)
{
    ConstantExpression constant = expression.Body as ConstantExpression;
    if (constant == null)
    {
        throw new ArgumentException("Expression must be a constant expression.");
    }
    UseNames(typeof(T).Name, constant.Value.ToString());
}

Use it like this:

UseNames<Foo>(x => nameof(x.Bar));
UseNames<Foo>(x => nameof(x.Baz));

edited Sep 25, 2019 at 22:51

Peter Mortensen

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answered Sep 25, 2019 at 15:33

GvS

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1 Comment

ckuri Over a year ago

Why the nameof? Your code seems to work with UseNames<Foo>(x => x.Bar).

Florian S. · Accepted Answer · 2019-09-25 18:10:45Z

1

A simple (and maybe faster) solution, without expression trees (which use reflection):

public string PrintFullName<T>(String memberName)
{
    return $"{typeof(T).Name}.{memberName}";
}

Usage:

PrintFullName<SomeType>(nameof(SomeType.SomeProperty));
// Compiled to: PrintFullName<SomeType>("SomeProperty");

nameof() is a compile-time construct, so no need to build a slightly advanced solution that performs evaluation at runtime more than required.

edited Sep 25, 2019 at 18:10

answered Sep 25, 2019 at 17:44

Florian S.

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