CODEWITHSUNDEEP
c#roslynroslyn-code-analysis
Görkem Özer
Görkem Özer

Reputation: 564

C# Roslyn Compiler - How to get namespace of a type from IdentifierNameSyntax?

Suppose I have a call in the code, SomeClass.SomeStaticMethod<T>(), which is an InvocationExpressionSyntax.

I get name of the generic type T as string (from IdentifierNameSyntax). I tried to get Symbol of T but I did not succeed.

How can I find out the namespace information of type T?

UPDATE: @SJP's answer is correct. I want to explain my mistake for those who want to get namespace information from IdentifierNameSyntax, which contains an identifier for a class (class name):

My initial aim was to find invocations in SomeClass.SomeMethod<T>() format and get namespace information of type T.

var namedTypeSymbol = context.Symbol as INamedTypeSymbol;
var reference = nameTypeSymbol.DeclaringSyntaxReferences.First();
var classSyntaxTree = reference.SyntaxTree;

var semanticModel = context.Compilation.GetSemanticModel(classSyntaxTree);
var genericNameSyntax = (GenericNameSyntax)((MemberAccessExpressionSyntax)node.Expression).Name;
var identifierNameSyntax = genericNameSyntax.TypeArgumentList.Arguments.First();
var typeInfo = semanticModel.GetTypeInfo(identifierNameSyntax);
var nameSpace = ((INamedTypeSymbol)typeInfo.Type).ContainingNamespace;
var nameSpaceName = nameSpace.ToString();

Here is my mistake:

I tried to get full namespace like <module_name>.<namespace_part_1>.<namespace_part_2> but when I did namedTypeSymbol.ContainingNamespace.Name , I got only <namespace_part_2>. After couple of hours, I found that getting full namespace is done like namedTypeSymbol.ContainingNamespace.ToString().

Sometimes best thing to do is going outside and taking fresh air :)

Upvotes: 13

Views: 5488

Answers (3)

Brandon Minnick
Brandon Minnick

Reputation: 15410

The ToDisplayString method lets you pass in a format object which has a number of options for controlling how you want to format stuff.

Passing SymbolDisplayTypeQualificationStyle.NameAndContainingTypesAndNamespaces will ensure that both the Namespace + Class name are returned.

var symbolDisplayFormat = new SymbolDisplayFormat(typeQualificationStyle: SymbolDisplayTypeQualificationStyle.NameAndContainingTypesAndNamespaces);

string fullyQualifiedName = typeSymbol.ToDisplayString(symbolDisplayFormat);

Upvotes: 4

jeff cassar
jeff cassar

Reputation: 3

A call to Function{SuppliedArgument}(...) is gonna have a GenericNameSyntax as the {SuppliedArgument}. Which isnt the same as a TypeDeclaration.TypeParameterList.Parameters. You can't get anything from those. Theyre placeholders. You need to go through the GenericNameSyntax.TypeArgumentList, and use SemanticModel.GetSymbolInfo(identifierNameSyntax).Symbol.ContainingNamespace on each to get the namespaces the identifiers are contained in.

Upvotes: 0

SJP
SJP

Reputation: 1010

You're going to need the semantic model to achieve your task. Assuming you require the Namespace of SomeClass, you can then just obtain the type and name from there the namespace from the MemberAccessExpressionSyntax by accessing the Name Field of the Expression as follows:

var semanticModel = await document.GetSemanticModelAsync()
var name = (GenericNameSyntax)((MemberAccessExpressionSyntax)node.Expression).Name;
var typeInfo = semanticModel.GetTypeInfo(name.TypeArgumentList.Arguments.First());
var nameSpace = ((INamedTypeSymbol)typeInfo.Type).ContainingNamespace;
var nameSpaceName = nameSpace.Name;

For the following example Program this would result in "System" or "ConsoleApp1" (depending on the call) for the variable nameSpaceName while all other information can be accessed by the variable nameSpace.

namespace ConsoleApp1
{
    class Program
    {
        static void Main(string[] args)
        {
            Program.DoStuff<string>();
            Program.DoStuff<Program>();
        }

        static void DoStuff<T>()
        {

        }
    }
}

Upvotes: 8

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