Fix code generation for Durable Functions function-based syntax. #47
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@cgillum I found another case to fully qualify the type name. |
| { | ||
| this.FullTypeName = fullTypeName; | ||
| this.RequiredNamespaces = requiredNamespaces; | ||
| this.Name = name; | ||
| this.Kind = kind; | ||
| this.Parameter = parameter; | ||
| this.ReturnType = returnTypeSyntax; | ||
| this.ReturnType = SyntaxNodeUtility.GetRenderedTypeExpression(returnType, false); |
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Does this false mean we don't support nullable annotations in return types?
| return false; | ||
| } | ||
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| if (!SyntaxNodeUtility.TryGetQualifiedTypeName(model, method, out string? fullTypeName)) | ||
| { | ||
| function = null; | ||
| return false; | ||
| } |
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Would it make sense to move the TryGetParameter and TryGetQualifiedName checks further up, above the new return type code you added? Could that short-circuiting help save some CPU cycles or do we very rarely expect those methods to return false?
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| return false; | ||
| } | ||
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| if (toProcess is not PredefinedTypeSyntax && typeInfo.Type.ContainingNamespace.IsGlobalNamespace) | ||
| if (typeInfo.ContainingNamespace.IsGlobalNamespace) | ||
| { | ||
| requiredNamespaces = null; | ||
| return false; |
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I know you didn't write this code, but is it correct for us to bail like this if we encounter a global namespace? Why wouldn't we continue to process remaining types in the remaining list?
| string expression = symbol.ToString(); | ||
| if (expression.StartsWith("System.") && symbol.ContainingNamespace.Name == "System") | ||
| { | ||
| expression = expression.Substring("System.".Length); |
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We do this because the System namespace is implicitly imported by the compiler?
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I have opened #48 to track the comments here and other improvements to the generator. |
This PR addresses 2 issues with function based syntax code gen for Durable Functions:
Task<T>would get double nested asTask<Task<T>>resolves #35