为什么在 Dart 的异步函数中没有以这种方式捕获异常？

Question

Future<void> fetchUserOrder() async {
// Imagine that this function is fetching user info but encounters a bug
  try {
  return Future.delayed(const Duration(seconds: 2),
      () => throw Exception('Logout failed: user ID is invalid'));
    
  } catch(e) {
    // Why exception is not caught here?
    print(e);
  }
}

void main() {
  fetchUserOrder();
  print('Fetching user order...');
}

输出

Fetching user order...
Uncaught Error: Exception: Logout failed: user ID is invalid

这表示未捕获异常。但是如您所见，throw Exception 子句被 try catch 包围。

Answer 1

try-catch 块只会捕获 awaited Future 的异常。所以你必须在你的代码中使用await：

Future<void> fetchUserOrder() async {
// Imagine that this function is fetching user info but encounters a bug
  try {
  return await Future.delayed(const Duration(seconds: 2),
      () => throw Exception('Logout failed: user ID is invalid'));
    
  } catch(e) {
    // Why exception is not caught here?
    print(e);
  }
}

void main() {
  fetchUserOrder();
  print('Fetching user order...');
}