Use value of property expression inside expression tree

Question

Consider a property expression like t => t.MyProperty where t is of type MyClass. How can I use this property expression inside a new expression where I perform a method call?

Pure C#:

class MyClass
{
    public string MyProperty { get; set; }
}

static void Foo(string foo)
{   
}

LambdaExpression GetExpression(Expression<Func<MyClass, object>> expr)
{
    return expr;
}

var myClass = new MyClass();
Foo(myClass.MyProperty);

Now with expressions...?

var expr = GetExpression(m => m.MyProperty);
var mi = typeof(Program).GetMethod(nameof(Program.Foo),
    BindingFlags.Public | BindingFlags.Static);

var myClass = new MyClass();
// Now what??
// var call = Expression.Call(mi, ???expr??);
// var invoke = Expression.Invoke(call, fooParameter);

I want to use the result of expr and use that in the call to Foo. I know I can do this in two steps, where I call expr.Compile().DynamicInvoke(myClass) to get the value, but that is not what I'm asking for here.

I want to build an expression that takes a property getter expression and then performs a call to Foo(result of expression). I cannot figure out how to use the expression as a parameter to the method call.

It's not clear at all what you are trying to achieve. Method Foo takes a string; your Expression produces an object. What expression would you like to use as a parameter to what method call? — Sergey Kalinichenko
– Sergey Kalinichenko, Commented Apr 7, 2017 at 14:57
if I understand it, you should just be able to use Expression.Call(mi, Expression.Invoke( ... )) ? — Marc Gravell
– Marc Gravell, Commented Apr 7, 2017 at 14:59
That's part of the challenge. Need for conversion, maybe. It is clear what I'm trying to achieve. For a given MyClass object M, I want to call Foo(M.MyProperty). The input is an expression as produced by GetExpression method above. — l33t
– l33t, Commented Apr 7, 2017 at 14:59
@l33t Foo takes a string, while Expression produces an object. Do you want a simple cast in between, because you know that the object is actually a string? — Sergey Kalinichenko
– Sergey Kalinichenko, Commented Apr 7, 2017 at 15:03

Marc Gravell · Accepted Answer · 2017-04-07 15:08:27Z

3

There's two ways of doing this, depending on the complexity. In this case, we can re-use the parameter from the inner expression - bubbling it outwards; we do this by discarding our old lambda, just using the .Body and .Parameters. For example:

var call = Expression.Lambda<Action<MyClass>>(
    Expression.Call(mi, expr.Body), expr.Parameters);

var myClass = new MyClass { MyProperty = "yay!" };
call.Compile().Invoke(myClass);

The other way is to use Invoke on the inner lambda:

var outerParameter = Expression.Parameter(typeof(MyClass));
var typed = Expression.Convert(Expression.Invoke(expr, outerParameter), typeof(string));
var call = Expression.Lambda<Action<MyClass>>(Expression.Call(mi, typed), outerParameter);

The second form (Invoke) is useful when you can't conveniently control the parameters in the two places - or where, for example, you have multiple inner expressions with different parameter instances.

edited Apr 7, 2017 at 15:08

answered Apr 7, 2017 at 15:04

Marc Gravell

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3 Comments

l33t Over a year ago

Wow. Saved my day. Thanks!

l33t Over a year ago

Second approach is actually very useful in my case. Thanks.

Marc Gravell Over a year ago

@l33t there's actually a third option if you want: use a custom ExpressionVisitor to replace all of one expression with another expression - for example, to substitute an inner parameter with an outer parameter, but works for any expression. Can provide examples if you want.

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