2 references to JsonExceptionTypeRef
System.Text.Json.SourceGeneration (2)
JsonSourceGenerator.Emitter.cs (2)
690writer.WriteLine($"_ => throw new {JsonExceptionTypeRef}(),"); 1636throw new {{JsonExceptionTypeRef}}(string.Format("{{ExceptionMessages.PropertyGetterDisallowNull}}", propertyName, {{FormatStringLiteral(typeGenSpec.TypeRef.Name)}}));