Swift：异步/等待超时 function

Question

如果时间太长，有没有办法附加超时以停止异步 function？

这是我的代码的简化版本：

func search() async -> ResultEnum {
  // Call multiple Apis one after the other
}

Task.init {
  let searchResult = await search() 
  switch searchResult {
    // Handle all result scenarios
  }
}

我希望search() async function 有一个提供结果的截止日期，否则它应该终止并返回ResultEnum.timeout 。

Answer 1

感谢 Rob 的评论以及您提供的链接。

不过，我不得不进行一些更改，出于某种原因，初始任务fetchTask即使在取消后仍继续运行，直到我向其添加Task.checkCancellation() 。

下面是代码现在的样子，如果有人遇到类似的问题：

func search() async throws -> ResultEnum {
  // This is the existing method as per my initial question.
  // It calls multiple Apis one after the other, then returns a result.
}

// Added the below method to introduce a deadline for search()
func search(withTimeoutSecs: Int) async {
  let fetchTask = Task {
        let taskResult = try await search()
        try Task.checkCancellation()
        // without the above line, search() kept going until server responded long after deadline.
        return taskResult
    }
    
    let timeoutTask = Task {
        try await Task.sleep(nanoseconds: UInt64(withTimeoutSecs) * NSEC_PER_SEC)
        fetchTask.cancel()
    }
    
    do {
        let result = try await fetchTask.value
        timeoutTask.cancel()
        return result
    } catch {
        return ResultEnum.failed(NetworkError.timeout)
    }
}

---

// Call site: Using the function (withTimeout:) instead of ()
Task.init {
  let searchResult = await search(withTimeoutSecs: 6) 
  switch searchResult {
    // Handle all result scenarios
  }
}