我试图在以下情况下可靠地比较ITypeSymbol最简单，最直接的两种情况（我在一个更大的项目中遇到了这些问题，并试图尽可能地简化它）：
我使用这个SyntaxTree获得了一个CSharpCompilation：
namespace MyAssembly
{
public class Foo
{
public Foo(Foo x)
{
}
}
}
我们正在使用CSharpSyntaxRewriter穿过树，更改班级并更新Compilation。在第一次运行中，我们记住第一个构造函数参数的ITypeSymbol（在这种情况下，这是类本身的类型）。
更新编译后，我们再次调用相同的重写器，并再次从构造函数参数中获取ITypeSymbol。
之后，我比较了我期望代表相同类型MyAssembly.Foo的两个ITypeSymbol。
我的第一个比较方法是调用ITypeSymbol.Equals()方法，但它返回false。它基本上返回false因为我们改变了编译并在此期间获得了新的SemanticModel。如果我们不这样做，Equals（）方法实际上返回true。
比较DeclaringSyntaxReferences（这里也说明How to compare type symbols (ITypeSymbol) from different projects in Roslyn?）会返回false，因为我们在此期间更改了类Foo。如果构造函数参数的类型为Bar，并且我们重写了Bar，则行为将是相同的。要验证这一点，只需取消注释行
//RewriteBar(rewriter, compilation, resultTree);
并在代码示例中用Bar替换构造函数参数类型。
结论：
ITypeSymbol.Equals()不适用于新的编译和语义模型，并且比较DeclaringSyntaxReferences不适用于我们在此期间更改的类型。
（我还使用一种外部程序集测试了行为 - 在这种情况下，ITypeSymbol.Equals（）为我工作。）
所以我的问题是：
这是完整的测试程序，使用该程序可以重现该问题。只需复制，包括Roslyn引用并执行：
using System;
using System.Collections.Generic;
using System.Linq;
using Microsoft.CodeAnalysis;
using Microsoft.CodeAnalysis.CSharp;
using Microsoft.CodeAnalysis.CSharp.Syntax;
namespace Demo.TypeSymbol
{
class Program
{
static void Main(string[] args)
{
var compilation = (CSharpCompilation) GetTestCompilation();
var rewriter = new Rewriter(changeSomething: true);
var tree = compilation.SyntaxTrees.First(); //first SyntaxTree is the one of class MyAssembly.Foo
rewriter.Model = compilation.GetSemanticModel (tree);
//first rewrite run
var resultTree = rewriter.Visit (tree.GetRoot()).SyntaxTree;
compilation = UpdateIfNecessary (compilation, rewriter, tree, resultTree);
rewriter.Model = compilation.GetSemanticModel (resultTree);
//just for demonstration; comment in to test behaviour when we are rewriting the class Bar -> in this case use Bar as constructor parameter in Foo
//RewriteBar(rewriter, compilation, resultTree);
//second rewrite run
rewriter.Visit (resultTree.GetRoot());
//now we want to compare the types...
Console.WriteLine(rewriter.ParameterTypeFirstRun);
Console.WriteLine(rewriter.ParameterTypeSecondRun);
//=> types are *not* equal
var typesAreEqual = rewriter.ParameterTypeFirstRun.Equals (rewriter.ParameterTypeSecondRun);
Console.WriteLine("typesAreEqual: " + typesAreEqual);
//=> syntax references are not equal
if(rewriter.ParameterTypeFirstRun.DeclaringSyntaxReferences.Any())
{
var syntaxReferencesAreEqual =
rewriter.ParameterTypeFirstRun.DeclaringSyntaxReferences.First()
.Equals(rewriter.ParameterTypeSecondRun.DeclaringSyntaxReferences.First());
Console.WriteLine("syntaxReferencesAreEqual: " + syntaxReferencesAreEqual);
}
//==> other options??
}
private static CSharpCompilation UpdateIfNecessary(CSharpCompilation compilation, Rewriter rewriter, SyntaxTree oldTree, SyntaxTree newTree)
{
if (oldTree != newTree)
{
//update compilation as the syntaxTree changed
compilation = compilation.ReplaceSyntaxTree(oldTree, newTree);
rewriter.Model = compilation.GetSemanticModel(newTree);
}
return compilation;
}
/// <summary>
/// rewrites the SyntaxTree of the class Bar, updates the compilation as well as the semantic model of the passed rewriter
/// </summary>
private static void RewriteBar(Rewriter rewriter, CSharpCompilation compilation, SyntaxTree firstSyntaxTree)
{
var otherRewriter = new Rewriter(true);
var otherTree = compilation.SyntaxTrees.Last();
otherRewriter.Model = compilation.GetSemanticModel(otherTree);
var otherResultTree = otherRewriter.Visit(otherTree.GetRoot()).SyntaxTree;
compilation = UpdateIfNecessary(compilation, otherRewriter, otherTree, otherResultTree);
rewriter.Model = compilation.GetSemanticModel(firstSyntaxTree);
}
public class Rewriter : CSharpSyntaxRewriter
{
public SemanticModel Model { get; set; }
private bool _firstRun = true;
private bool _changeSomething;
public ITypeSymbol ParameterTypeFirstRun { get; set; }
public ITypeSymbol ParameterTypeSecondRun { get; set; }
public Rewriter (bool changeSomething)
{
_changeSomething = changeSomething;
}
public override SyntaxNode VisitClassDeclaration(ClassDeclarationSyntax node)
{
node = (ClassDeclarationSyntax)base.VisitClassDeclaration(node);
//remember the types of the parameter
if (_firstRun)
ParameterTypeFirstRun = GetTypeSymbol (node);
else
ParameterTypeSecondRun = GetTypeSymbol (node);
_firstRun = false;
//change something and return updated node
if(_changeSomething)
node = node.WithMembers(node.Members.Add(GetMethod()));
return node;
}
/// <summary>
/// Gets the type of the first parameter of the first method
/// </summary>
private ITypeSymbol GetTypeSymbol(ClassDeclarationSyntax classDeclaration)
{
var members = classDeclaration.Members;
var methodSymbol = (IMethodSymbol) Model.GetDeclaredSymbol(members[0]);
return methodSymbol.Parameters[0].Type;
}
private MethodDeclarationSyntax GetMethod()
{
return (MethodDeclarationSyntax)
CSharpSyntaxTree.ParseText (@"public void SomeMethod(){ }").GetRoot().ChildNodes().First();
}
}
private static SyntaxTree[] GetTrees()
{
var treeList = new List<SyntaxTree>();
treeList.Add(CSharpSyntaxTree.ParseText(Source.Foo));
treeList.Add(CSharpSyntaxTree.ParseText(Source.Bar));
return treeList.ToArray();
}
private static Compilation GetTestCompilation()
{
var mscorlib = MetadataReference.CreateFromFile(typeof(object).Assembly.Location);
var refs = new List<PortableExecutableReference> { mscorlib };
// I used this to test it with a reference to an external assembly
// var testAssembly = MetadataReference.CreateFromFile(@"../../../Demo.TypeSymbol.TestAssembly/bin/Debug/Demo.TypeSymbol.TestAssembly.dll");
// refs.Add (testAssembly);
return CSharpCompilation.Create("dummyAssembly", GetTrees(), refs);
}
}
public static class Source
{
public static string Foo => @"
// for test with external assembly
//using Demo.TypeSymbol.TestAssembly;
namespace MyAssembly
{
public class Foo
{
public Foo(Foo x)
{
}
}
}
";
public static string Bar => @"
namespace MyAssembly
{
public class Bar
{
public Bar(int i)
{
}
}
}
";
}
}
答案 0 :(得分：4)
一种可能性是调用SymbolFinder.FindSimilarSymbols，它会在新解决方案中为您提供符号名称和其他一些属性的符号。从那里你可以在你的新编辑中等于。