2013-03-24 99 views
1

我正在C#中编写一个连接语言,目前它被解释,但我想要采取下一步:编译。首先,我尝试写一个简单的“你好,世界!”程序发射器使用System.Reflection.Emit。该代码工作没有任何的Emit异常,但是当我运行生成的“TEST.EXE”的文件,它抛出我曾试着用搜索引擎寻找答案异常Reflection.Emit:AssemblyBuilder.SetEntryPoint不设置入口点

Unhandled Exception: System.MissingMethodException: Entry point not found in assembly 'IL_Test, Version=0.0.0.0, Culture=neutral, PublicKeyToken=null'. 

,但无济于事。也许这里有人可以帮助我? (请)我写的代码如下:

using System; 
using System.Collections.Generic; 
using System.Linq; 
using System.Text; 
using System.Reflection; 
using System.Reflection.Emit; 
using System.IO; 
using System.Diagnostics; 

namespace ILCompileTest 
{ 
    class Program 
    { 
     static void Main(string[] args) 
     { 
      const string ASSEMBLY_NAME = "IL_Test"; 

      AssemblyBuilder assemblyBuilder = AppDomain.CurrentDomain.DefineDynamicAssembly(
       new AssemblyName(ASSEMBLY_NAME), AssemblyBuilderAccess.Save); 
      ModuleBuilder moduleBuilder = assemblyBuilder.DefineDynamicModule(
       ASSEMBLY_NAME); 
      TypeBuilder typeBuilder = moduleBuilder.DefineType("Program", 
       TypeAttributes.Class | TypeAttributes.Public); 
      MethodBuilder methodBuilder = typeBuilder.DefineMethod(
       "Main", MethodAttributes.Public | MethodAttributes.Static, 
       typeof(void), new Type[] { typeof(string[]) }); 
      ILGenerator gen = methodBuilder.GetILGenerator(); 

      gen.Emit(OpCodes.Ldstr, "Hello, World!"); 
      gen.Emit(OpCodes.Call, typeof(Console).GetMethod("WriteLine", new Type[] { typeof(string) })); 
      gen.Emit(OpCodes.Ldc_I4_1); 
      gen.Emit(OpCodes.Call, typeof(Console).GetMethod("ReadKey", new Type[] { typeof(bool) })); 

      assemblyBuilder.SetEntryPoint(methodBuilder, PEFileKinds.ConsoleApplication); 
      File.Delete("test.exe"); 
      assemblyBuilder.Save("test.exe"); 

      Process.Start("test.exe"); 
     } 
    } 
} 

所以,问题是:我怎么可以在进入点设置为Main方法我定义?

回答

1

您正在调用typeBuilder.CreateType(), DefineDynamicModule必须将exe名称作为第二个参数传递。 Full working sample:

using System; 
using System.Collections.Generic; 
using System.Linq; 
using System.Text; 
using System.Reflection; 
using System.Reflection.Emit; 
using System.IO; 
using System.Diagnostics; 

namespace ILCompileTest 
{ 
    class Program 
    { 
     static void Main(string[] args) 
     { 
      const string ASSEMBLY_NAME = "IL_Test"; 

      AssemblyBuilder assemblyBuilder = AppDomain.CurrentDomain.DefineDynamicAssembly(
       new AssemblyName(ASSEMBLY_NAME), AssemblyBuilderAccess.Save); 
      ModuleBuilder moduleBuilder = assemblyBuilder.DefineDynamicModule(
       ASSEMBLY_NAME, "test.exe"); 
      TypeBuilder typeBuilder = moduleBuilder.DefineType("Program", 
       TypeAttributes.Class | TypeAttributes.Public); 
      MethodBuilder methodBuilder = typeBuilder.DefineMethod(
       "Main", MethodAttributes.HideBySig|MethodAttributes.Public | MethodAttributes.Static, 
       typeof(void), new Type[] { typeof(string[]) }); 
      ILGenerator gen = methodBuilder.GetILGenerator(); 

      gen.Emit(OpCodes.Ldstr, "Hello, World!"); 
      gen.Emit(OpCodes.Call, typeof(Console).GetMethod("WriteLine", new Type[] { typeof(string) })); 
      gen.Emit(OpCodes.Ldc_I4_1); 
      gen.Emit(OpCodes.Call, typeof(Console).GetMethod("ReadKey", new Type[] { typeof(bool) })); 
      typeBuilder.CreateType(); 
      assemblyBuilder.SetEntryPoint(methodBuilder, PEFileKinds.ConsoleApplication); 
      File.Delete("test.exe"); 
      assemblyBuilder.Save("test.exe"); 

      Process.Start("test.exe"); 
     } 
    } 
} 
+0

Thanks!这解决了它!我还发现,我需要Opcodes.Pop Console.ReadKey的返回值,并有一个Opcodes.Ret命令,以使整个事情工作。 – feralin 2013-03-24 19:41:28

+0

你知道为什么ModuleBuilder想要文件名吗?它甚至必须与我传递给AssemblyBuilder.Save()的值相同吗? – feralin 2013-03-24 19:42:22

+0

我认为这与DefineDynamicModule的重载选择差有关 - 根据MSDN,两个重载创建瞬态,两个创建持久模块。文件名必须相同,否则应用程序也会失败。在两个地方指定文件名可以从多个模块/文件中创建程序集。 – Alexander 2013-03-25 07:00:41