Why INamedTypeSymbol.TypeArguments all is INamedTypeSymbol? #80396
Replies: 3 comments 5 replies
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If you want the type parameters, then use the TypeParameters property. The TypeArguments property contains the substituted arguments for each parameter. |
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@CyrusNajmabadi Hi, But if I want to publish it as nuget so that people can directly reference it without adding |
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You can create a msbuild target file to include your source generator, so that people reference your package will get your source generator referenced automatically. |
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the code, T in TypeArguments ,it is IErrorTypeSymbol & INamedTypeSymbol . not is ITypeParameterSymbol.
Is this correct?
i want ITypeParameterSymbol
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