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Issue 11293132: Add a Futures.forEach function for asynchronously iterating over a collection. (Closed) Base URL: https://dart.googlecode.com/svn/branches/bleeding_edge/dart
Patch Set: Code review change Created 8 years, 1 month ago
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1 // Copyright (c) 2012, the Dart project authors. Please see the AUTHORS file 1 // Copyright (c) 2012, the Dart project authors. Please see the AUTHORS file
2 // for details. All rights reserved. Use of this source code is governed by a 2 // for details. All rights reserved. Use of this source code is governed by a
3 // BSD-style license that can be found in the LICENSE file. 3 // BSD-style license that can be found in the LICENSE file.
4 4
5 /** 5 /**
6 * A [Future] is used to obtain a value sometime in the future. Receivers of a 6 * A [Future] is used to obtain a value sometime in the future. Receivers of a
7 * [Future] can obtain the value by passing a callback to [then]. For example: 7 * [Future] can obtain the value by passing a callback to [then]. For example:
8 * 8 *
9 * Future<int> future = getFutureFromSomewhere(); 9 * Future<int> future = getFutureFromSomewhere();
10 * future.then((value) { 10 * future.then((value) {
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252 }); 252 });
253 future.handleException((exception) { 253 future.handleException((exception) {
254 if (!result.isComplete) { 254 if (!result.isComplete) {
255 completer.completeException(exception, future.stackTrace); 255 completer.completeException(exception, future.stackTrace);
256 } 256 }
257 return true; 257 return true;
258 }); 258 });
259 } 259 }
260 return result; 260 return result;
261 } 261 }
262
263 /**
264 * Runs [f] for each element in [input] in order, moving to the next element
265 * only when the [Future] returned by [f] completes. Returns a [Future] that
266 * completes when all elements have been processed.
267 *
268 * The return values of all [Future]s are discarded. Any errors will cause the
269 * iteration to stop and will be piped through the returned [Future].
270 */
271 static Future forEach(Iterable input, Future f(element)) {
272 var iterator = input.iterator();
273 Future nextElement(_) {
274 if (!iterator.hasNext) return new Future.immediate(null);
275 return f(iterator.next()).chain(nextElement);
276 }
277 return nextElement(null);
278 }
262 } 279 }
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