Roslyn可以从对象实例生成源代码吗？

问题描述

在Visual Studio 2015中使用Roslyn API，我可以将对象实例转换为源代码吗？是否可以创建如下所示的扩展方法，例如 .ToSourceCode（）？

Using the Roslyn API with Visual Studio 2015, can I convert an object instance to source code? Can I create an extension method like ".ToSourceCode()" as shown below?

class Foo { }
class Program
{
    static string classSourceCode = "class Foo { }";
    static void Main()
    {
        var instance = new Foo();
        var instanceSourceCode = instance.GetType().ToSourceCode();
        System.Diagnostics.Debug.Assert(instanceSourceCode == classSourceCode);
    }
}

推荐答案

否。但是，ILSpy可以。

No. However, ILSpy can.

基于对问题的评论以及我对Roslyn的了解，不支持反编译。但是，感谢@Bradley的ILSpy提示，有一个解决方案：

Based on the comments on the question and what I understand about Roslyn, decompilation is not supported. However, thanks to @Bradley's ILSpy tip, there is a solution:

1. 从


2. 引用以下程序集：ICSharpCode.Decompiler.dll，ILSpy.exe，Mono.Cecil.dll， ILSpy.BamlDecompiler.Plugin.dll


3. 实施 .ToSourceCode（）扩展方法，如下所示：

using System;
using System.Linq;
using System.Reflection;
using System.Text.RegularExpressions;
using ICSharpCode.Decompiler;
using ICSharpCode.ILSpy;
using Mono.Cecil;
class Foo { }
class Program
{
    static string classSourceCode = "using System; internal class Foo { } ";
    static void Main()
    {
        var instance = new Foo();
        var instanceSourceCode = instance.GetType().ToSourceCode();
        System.Diagnostics.Debug.Assert(instanceSourceCode == classSourceCode);
    }
}

static class TypeExtensions
{
    public static string ToSourceCode(this Type source)
    {
        var assembly = AssemblyDefinition.ReadAssembly(Assembly.GetExecutingAssembly().Location);
        var type = assembly.MainModule.Types.FirstOrDefault(t => t.FullName == source.FullName);
        if (type == null) return string.Empty;
        var plainTextOutput = new PlainTextOutput();
        var decompiler = new CSharpLanguage();
        decompiler.DecompileType(type, plainTextOutput, new DecompilationOptions());
        return Regex.Replace(Regex.Replace(plainTextOutput.ToString(), @"\n|\r", " "), @"\s+", " ");
    }
}

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