如何使用Roslyn在C＃中获得类型的lambda？

Question

我正在用罗斯林（Roslyn）编写我的代码缀合物。我希望它以这种方式工作：
以前：

IEnumerable<int> list = new List<int> {1, 2, 3, 4};
list = list.Where(x => x%2 == 0);

问题：

...
IEnumerable<int> list = new List<int> {1, 2, 3, 4};
list = list.Where(lambdaMethod);
...
private static bool lamdaMethod(int x)
{
    return x % 2 == 0;
}

问题是我找不到适当的方法和参数类型的返回类型。
我从分析仪模板中的方法：

public sealed override async Task RegisterCodeFixesAsync(CodeFixContext context)
        {
            var root = await context.Document.GetSyntaxRootAsync(context.CancellationToken).ConfigureAwait(false);

            // TODO: Replace the following code with your own analysis, generating a CodeAction for each fix to suggest
            var diagnostic = context.Diagnostics.FirstOrDefault();
            var diagnosticSpan = diagnostic.Location.SourceSpan;

            // Find the type declaration identified by the diagnostic.
            var declaration = root.FindToken(diagnosticSpan.Start).Parent.AncestorsAndSelf()
                .OfType<LambdaExpressionSyntax>().FirstOrDefault();
            if (declaration == null)
                return;
            // Register a code action that will invoke the fix.

            context.RegisterCodeFix(
                CodeAction.Create(
                    title: $"{"Just Title"}",
                    createChangedDocument: c => MakeLambdaOptimization(context.Document, declaration, c),
                    equivalenceKey: nameof(CodeFixResources.CodeFixTitle)),
                diagnostic);
        }

如您所见，我正在使用lambdaexpressionsyntax。还有SimpleLambdaexpressyNtax和括号>括号>括号lambdaexpressionsyntax，可以帮助我获得parametersyntax。在parametersyntax中，我可以找到type，但它是null，我得到了NullReferenceException。
我试图获得全部类型的lambda（例如func＆lt; int，bool＆gt;），我想这对我有帮助。我该怎么做？我听说过semanticmodel，但我不知道如何使用它

Answer 1

您可以使用 SemanticModel.GetSymbolInfo(ExpressionSyntax) 获取 lambda 符号。从 lambda 符号中您可以找到所需的信息（返回类型和参数类型）。这是一个代码片段（为了简单起见，在控制台应用程序而不是代码修复程序的上下文中编写 - 您将能够使其适用于代码修复程序）：

using Microsoft.CodeAnalysis;
using Microsoft.CodeAnalysis.CSharp;
using Microsoft.CodeAnalysis.CSharp.Syntax;

// You don't need to create or get a compilation yourself. I'm creating it manually since I'm in a console app.
CSharpCompilation comp = CSharpCompilation.Create(null, new[] { SyntaxFactory.ParseSyntaxTree("System.Func<int, bool> f = x => x % 2 == 0;") }, new[] { MetadataReference.CreateFromFile(typeof(Func<>).Assembly.Location) });

// You already have the LambdaExpressionSyntax, ignore the following two lines, which again because I'm in a console app.
var root = comp.SyntaxTrees[0].GetRoot();
var lambda = root.DescendantNodes().OfType<LambdaExpressionSyntax>().Single();

// Here you'll need to get a semantic model via 'await context.Document.GetSemanticModelAsync(context.CancellationToken).ConfigureAwait(false);'
var model = comp.GetSemanticModel(lambda.SyntaxTree);

// That's the core of what you want to do.
if (model.GetSymbolInfo(lambda).Symbol is IMethodSymbol lambdaSymbol)
{
    // prints 'bool'
    Console.WriteLine(lambdaSymbol.ReturnType.ToDisplayString());
    foreach (var parameter in lambdaSymbol.Parameters)
    {
        // prints 'int' (loop is entered once since my lambda has a single parameter).
        Console.WriteLine(parameter.Type.ToDisplayString());
    }
}