DART: syntax of future then

2019-06-24 07:52发布

问题:

I don´t understand the syntax of the then() clause.

1. myFuture(6).then( (erg) => print(erg) )

What´s (erg) => expr syntactically?

I thougt it could be a function, but

then( callHandler2(erg)

doesn´t work, Error:

"Multiple markers at this line
- The argument type 'void' cannot be assigned to the parameter type '(String) -> 
 dynamic'
- Undefined name 'erg'
- Expected to find ')'"

2. myFuture(5).then( (erg) { callHandler(erg);}, onError: (e) => print (e)

What´s `onError: (e) => expr"` syntactically?

3. Is there a difference between the onError: and the .catchError(e) variants?

回答1:

1) The Fat Arrow is syntactic sugar for short anonymous functions. The two functions below are the same:

someFuture(arg).then((erg) => print(erg));
// is the same as
someFuture(arg).then((erg) { return print(erg); });

Basically the fat arrow basically automatically returns the evaluation of the next expression.

If your callHandler2 has the correct signature, you can just pass the function name. The signature being that it accept the number of parameters as the future will pass to the then clause, and returns null/void.

For instance the following will work:

void callHandler2(someArg) { ... }
// .. elsewhere in the code
someFuture(arg).then(callHandler);

2) See answer 1). The fat arrow is just syntactic sugar equivalent to:

myFuture(5).then( (erg){ callHandler(erg);}, onError: (e){ print(e); });

3) catchError allows you to chain the error handling after a series of futures. First its important to understand that then calls can be chained, so a then call which returns a Future can be chained to another then call. The catchError will catch errors both synchronous and asynchronous from all Futures in the chain. Passing an onError argument will only deal with an error in the Future its an argument for and for any synchronous code in your then block. Any asynchronous code in your then block will remain uncaught.

Recent tendency in most Dart code is to use catchError and omit the onError argument.