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...');
}
它輸出
Fetching user order...
Uncaught Error: Exception: Logout failed: user ID is invalid
這表示未捕獲例外。但是如您所見，該throw Exception子句被 try catch 包圍。
uj5u.com熱心網友回復：
try-catch 塊只會捕獲等待的Future 的例外。所以你必須await在你的代碼中使用：
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...');
}
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