CSharpCodeProvider符合接口

Question

我有一个CSharpCodeProvider，它接受一个类的代码。在编译代码的项目中，我有一个接口。我希望我正在编译的代码符合这个接口。

这是我能想到的最简单的例子来说明我的问题。我有一个包含两个文件的项目：

Program.cs：

namespace ConsoleApplication1
{
    class Program
    {
        static void Main(string[] args)
        {
            //set up the compiler
            CSharpCodeProvider csCompiler = new CSharpCodeProvider();

            CompilerParameters compilerParameters = new CompilerParameters();
            compilerParameters.GenerateInMemory = true;
            compilerParameters.GenerateExecutable = false;

            var definition = 
@"class Dog : IDog
{
    public void Bark()
    {
        //woof 
    }
}";    
            CompilerResults results = csCompiler.CompileAssemblyFromSource(compilerParameters, new string[1] { definition });

            IDog dog = null;
            if (results.Errors.Count == 0)
            {
                Assembly assembly = results.CompiledAssembly;
                dog = assembly.CreateInstance("TwoDimensionalCellularAutomatonDelegate") as IDog;
            }

            if (dog == null)
                dog.Bark();
        }
    }
}

IDog.cs：

namespace ConsoleApplication1
{
    interface IDog
    {
        void Bark();
    }
}

我似乎想不出如何让CSharpCodeProvider识别IDog。我尝试过compilerParameters.ReferencedAssemblies.Add("ConsoleApplication1");，但它不起作用。任何帮助都将不胜感激。

Answer 1

解决方案是引用当前正在执行的程序集。

var location = Assembly.GetExecutingAssembly().Location;
compilerParameters.ReferencedAssemblies.Add(location);

Answer 2

尝尝这个

using System;
using System.IO;
using System.CodeDom.Compiler;
using System.Reflection;
using Microsoft.CSharp;

namespace ConsoleApplication1
{
    class Program
    {
        static void Main(string[] args)
        {
            //set up the compiler
            CSharpCodeProvider csCompiler = new CSharpCodeProvider();

            CompilerParameters compilerParameters = new CompilerParameters();
            compilerParameters.GenerateInMemory = true;
            compilerParameters.GenerateExecutable = false;
            compilerParameters.ReferencedAssemblies.Add("System.dll");

            var definition = File.ReadAllText("./IDog.cs") +
@"public class Dog : ConsoleApplication1.IDog
{
    public void Bark()
    {
        System.Console.WriteLine(""BowWoW"");
    }
}";
            CompilerResults results = csCompiler.CompileAssemblyFromSource(compilerParameters,
                new string[1] { definition });

            dynamic dog = null;
            if (results.Errors.Count == 0)
            {
                Assembly assembly = results.CompiledAssembly;
//              Type idog = assembly.GetType("ConsoleApplication1.IDog");
                dog = assembly.CreateInstance("Dog");
            } else {
                Console.WriteLine("Has Errors");
                foreach(CompilerError err in results.Errors){
                    Console.WriteLine(err.ErrorText);
                }
            }
            if (dog != null){
                dog.Bark();
            } else {
                System.Console.WriteLine("null");
            }
        }
    }
}

Answer 3

太棒了！可能会吧

IDog dog = (IDog)a.CreateInstance("Dog");
dog.Bark(/*additional parameters go here*/);