Task{} 有效，但 .task{} 抛出异常 抛出类型错误的函数导致无效转换？

Question

我试图理解为什么它Task{} 有效的原因，但是当我这样做时，VStack{}.task{}我收到错误：

从“@Sendable () async throws -> ()”类型的抛出函数到非抛出函数类型“@Sendable () async -> Void”的转换无效

我试图让该fetchWeather()函数在视图启动时运行，而无需用户点击按钮。（如果下面的方法是正确的方法）但是遇到了这个错误并且我真的很好奇这个错误背后的原因

下面是我的代码：

struct WeatherView: View {
    var body: some View {
        VStack{
           Text("Hello, World!")
           Button("Get Weather", action: {
            // Works when I tap the button
//                Task {
//                    try await fetchWeather()
//                }
        })
    }
    //*******Xcode error with invalid conversion*******
    .task {
        try await fetchWeather()
    }
  }
}

struct WeatherView_Previews: PreviewProvider {
   static var previews: some View {
       WeatherView( )
   }
}

功能：

func fetchWeather () async throws  {
   let URLString = "https://api.openweathermap.org/data/2.5/weather?appid=someAPIKeyshere&q=seattle"
   let (data, response) = try await URLSession.shared.data(from:  URL(string: URLString)!)
   guard let httpResponse = response as? HTTPURLResponse,
      httpResponse.statusCode == 200 else {
          throw ResponseError.someError
      }
   let decodedResponse = try? JSONDecoder().decode(WeatherData.self, from: data)
   print("This is in decodedResponse: \(decodedResponse)")
}

提前致谢！

Answer 1

查看您收到的错误消息\xe2\x80\xa6

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您的函数类型为：

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@Sendable () async throws -> ()\n

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而.task修饰符期望\xe2\x80\xa6

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@Sendable () async -> Void\n

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区别在于throws.

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要解决这个问题，您需要处理函数抛出的错误，例如\xe2\x80\xa6

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.task {\n    do {\n        try await fetchWeather()\n    } catch {\n        // handle error\n    }\n}\n

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您可以比较初始化程序Task和.task修饰符采用的参数：

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Task.init

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init(priority: TaskPriority? = nil, operation: @escaping @Sendable () async throws -> Success)\n

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.task修饰语

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func task(priority: TaskPriority = .userInitiated, _ action: @escaping @Sendable () async -> Void) -> some View\n

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