johnniwinther@google.com
cb7bedf860
The overall strategy is to convert the following:
switch (e) {
l_0: case e0: s_0; break;
l_1: case e1: s_1; continue l_i;
...
l_n: default: s_n; continue l_j;
}
into this:
var label;
switch (e) {
case e0: label = 0; break;
case e1: label = 1; break;
...
default: label = n; break;
}
l: while (true) {
switch (label) {
case 0: s_0; break l;
case 1: s_1; label = i; continue l;
...
case n: s_n; label = j; continue l;
}
}
BUG=http://dartbug.com/8270
R=ngeoffray@google.com
Review URL: https://codereview.chromium.org//14969004
git-svn-id: https://dart.googlecode.com/svn/branches/bleeding_edge/dart@22838 260f80e4-7a28-3924-810f-c04153c831b5
42 lines
998 B
Dart
42 lines
998 B
Dart
// Copyright (c) 2013, the Dart project authors. Please see the AUTHORS file
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// for details. All rights reserved. Use of this source code is governed by a
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// BSD-style license that can be found in the LICENSE file.
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// Test nested switch statement using labels.
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library nested_switch_label;
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import "package:expect/expect.dart";
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void main() {
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doSwitch(0, ['0', '2:0', '1', 'default']);
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doSwitch(2, ['2:2', '2:1', '2', '1', 'default']);
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}
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void doSwitch(int target, List expect) {
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List list = [];
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switch (target) {
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outer0: case 0:
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list.add('0');
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continue outer2;
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outer1: case 1:
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list.add('1');
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continue outerDefault;
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outer2: case 2:
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switch (target) {
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inner0: case 0:
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list.add('2:0');
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continue outer1;
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inner2: case 2:
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list.add('2:2');
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continue inner1;
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inner1: case 1:
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list.add('2:1');
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}
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list.add('2');
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continue outer1;
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outerDefault: default:
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list.add('default');
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}
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Expect.listEquals(expect, list);
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} |