想要推遲某些代碼行，使用 await Future 并且效果很好，問題是它推遲了它之后的所有代碼，我需要它推遲某些代碼行，同時繼續立即執行其余代碼
void main() async {
for (int i = 0; i < 5; i ) {
await Future.delayed(Duration(seconds: 1));
//postpone just next line or few lines of code
print('postpone this line of code ${i 1}');
print('postpone me too');
}
//should execute without being postponed
print('continue imediately without being postponed by await Future');
}
這可以通過 await Future 或其他一些功能實作嗎？
uj5u.com熱心網友回復：
await它是注冊Future.then回呼的語法糖。使用的目的await是更容易讓所有后續代碼等待Future完成。如果這不是你想要的，你可以Future.then直接使用：
void main() {
for (int i = 0; i < 5; i ) {
Future.delayed(Duration(seconds: 1)).then((_) {
print('postpone this line of code ${i 1}');
print('postpone me too');
});
}
print('continue immediately without being postponed by await Future');
}
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