I have a method createClient which return Future value and this method call another _getConfig method which returns a response as a Function callback, is there any way that in createClient method I can return Future from response of _getConfig method.
Future<String> createClient() async {
_getSocketConfig((socketConfig) => Future.value(
socketConfig,
));
}
void _getConfig(Function(String) onConfigFetch) async {
onConfigFetch("profileId");
}
I'm not sure if I understand, but is it this what you want?
Future<String> createClient() async {
return _getSocketConfig((socketConfig) => Future.value(
socketConfig,
));
}
Future<String> _getSocketConfig(Function(String) onConfigFetch) async {
return onConfigFetch("profileId");
}
Related
main() async {
try {
final t = Test();
await Future.delayed(Duration(seconds: 1));
} catch (e) {
// Never printed
print("caught");
}
}
void willThrow() async {
throw "error";
}
class Test {
Test() {
willThrow();
}
}
If the "async" keyword is removed from willThrow everything works as expected.
Is it because you can't await a constructor? If so is there anyway to catch async errors in a constructor body?
Have this a go:
void main() async {
try {
final t = Test();
await Future.delayed(Duration(seconds: 1));
} catch (e) {
// Never printed
print("caught");
}
}
Future<void> willThrow() async {
throw "error";
}
class Test {
Test() {
willThrow().catchError((e){print('Error is caught here with msg: $e');});
}
}
As to the 'why':
You use a normal try/catch to catch the failures of awaited asynchronous computations. But since you cannot await the constructor, you have to register the callback that handles the exception in another way. I think :)
Since you never awaited the Future that was returned from willThrow(), and you never used the result of the Future, any exception thrown by the function is discarded.
There is no way to write an asynchronous constructor. So you are stuck with using old-school callbacks to handle errors, or simulate an async constructor with a static method:
void main() async {
try {
final t = await Test.create();
await Future.delayed(Duration(seconds: 1));
} catch (e) {
// Never printed
print("caught");
}
}
Future<void> willThrow() async {
throw "error";
}
class Test {
Test._syncCreate() {}
Future<void> _init() async {
await willThrow();
}
static Test create() async {
Test result = Test._syncCreate();
await result._init();
return result;
}
}
Can some one help me to identify the issue in below piece of code
void main() async {
bool c =getstatus();
print(c);
}
Future<bool> getMockData() {
return Future.value(false);
}
bool getstatus() async
{
Future<bool> stringFuture = getMockData();
bool message = stringFuture;
return(message); // will print one on console.
}
To get values from a Future(async) method, you have to await them. And after await the variable you get is not a Future anymore. So basically your code should look like this:
void main() async {
bool c = await getstatus();
print(c);
}
Future<bool> getMockData() {
return Future.value(false);
}
Future<bool> getstatus() async {
bool message = await getMockData();
return message;
}
Future<bool> stringFuture = await getMockData();
an async method must return Future of something then in the main you have to get the bool value by writing await
void main() async {
bool c = await getstatus();
print(c); // will print false on the console.
}
Future<bool> getMockData() {
return Future.value(false);
}
Future<bool> getstatus() async {
Future<bool> stringFuture = await getMockData();
return stringFuture; // will return false.
}
I have a function A() which return a string and I need to receive the string and print the string using then() method in dart
Future<String> A() async {
return "Hello";
}
await A().then((value) => print(value));
I want to get the "Hello" printed but it prints null. How to do this ?
await should be used in an async function. But you have used await outside of the async function.
Try removing await outside.
Future<String> A() async {
return "Hello";
}
A().then((value) => print(value));
How can i return a string from a future function?
Future<String> functionA() async {
var x = await fetchX();
return x;
}
Future<String> fetchX() {
return Future.delayed(Duration(seconds: 4), () => 'example');
}
Future<String> la() async {
print(await functionA()); //this works correctly
return await functionA(); //this return always an instance of Future
}
How can i return "example" from the future function, there is a method to do it, and where is my error?
Future<String> fetch() async {
return
http.get('url')
.then((response) => response.body);
}
That way you can sneak a .catchError into there. :)
You need to specify what your function will return. All you have to do is add Future to the beginning of the method.
Future<String> fetch() async {
final response = await http.get('url');
String conteggio = response.body;
return conteggio;
}
And you have to do this in a method. You can only assign constant values in fields other than methods.
I'm trying to pass as parameter of a method, a function that returns Future<Response>.
I tried to do
Future<String> _execute(Function<Future<Response>>() function) async { }
but it does not even compile.
What's the correct syntax?
You can do it like this,
Future<String> _myFunction(Future<Response> Function() function) {
...
}
You just need to specify that your parameter is a Function:
Future<bool> kappa() async{
await Future.delayed(Duration(seconds: 1));
return true;
}
Future<bool> foo(Function f) async{
var k = await f();
return k;
}
void main() async{
print(await foo(kappa));
}
This will print true. In your case, your function parameter can be:
Future<String> _execute(Function function) async { }