[Solved] What is the issue with this Swift async GET request code?func fetchData() async { let url = URL(string: "https://example.com")! let (data, response) = URLSession.shared.data(from: url) print(data) } | iOS Swift

iOS Swift - Networking

What is the issue with this Swift async GET request code?

func fetchData() async {
  let url = URL(string: "https://example.com")!
  let (data, response) = URLSession.shared.data(from: url)
  print(data)
}

AMissing both 'try' and 'await' keywords before the data call

BURL is not unwrapped safely

CThe function should not be marked async

DThe print statement should be inside a do-catch block

Step-by-Step Solution

Solution:

Step 1: Identify async call requirements
The method URLSession.shared.data(from:) is asynchronous and throwable, so it requires try and await.
Step 2: Check the code
The code calls the method without try and await, which is incorrect.
Final Answer:
Missing both 'try' and 'await' keywords before the data call -> Option A
Quick Check:
Async throwable calls need both try and await [OK]

Quick Trick: Async throwable calls require try and await keywords [OK]

Common Mistakes:

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What is the issue with this Swift async GET request code?