Access the value of a member expression

You can compile and invoke a lambda expression whose body is the member access:

private object GetValue(MemberExpression member)
{
    var objectMember = Expression.Convert(member, typeof(object));

    var getterLambda = Expression.Lambda<Func<object>>(objectMember);

    var getter = getterLambda.Compile();

    return getter();
}

Local evaluation is a common technique when parsing expression trees. LINQ to SQL does this exact thing in quite a few places.

 MemberExpression right = (MemberExpression)((BinaryExpression)p.Body).Right;
 Expression.Lambda(right).Compile().DynamicInvoke();

The constant expression is going to point to a capture-class generated by the compiler. I've not included the decision points etc, but here's how to get 30 from that:

var p = new Product { Price = 30 };
Expression<Func<Product, bool>> predicate = x => x.Price == p.Price;
BinaryExpression eq = (BinaryExpression)predicate.Body;
MemberExpression productToPrice = (MemberExpression)eq.Right;
MemberExpression captureToProduct = (MemberExpression)productToPrice.Expression;
ConstantExpression captureConst = (ConstantExpression)captureToProduct.Expression;
object product = ((FieldInfo)captureToProduct.Member).GetValue(captureConst.Value);
object price = ((PropertyInfo)productToPrice.Member).GetValue(product, null);

price is now 30. Note that I'm assuming that Price is a property, but in reality you would write a GetValue method that handles property / field.

If you had a class:

public class Item
{
    public int Id { get; set; }
}

and an instance of the object:

var myItem = new Item { Id = 7 };

You can get the value of Id using an Expression using the following code:

Expression<Func<Item, int>> exp = x => x.Id;
var me = exp.Body as MemberExpression;
var propInfo = me.Member as PropertyInfo;
var myValue = propInfo.GetValue(myItem, null);

myValue will contain "7"