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// Copyright 2013 the V8 project authors. All rights reserved.
// Redistribution and use in source and binary forms, with or without
// modification, are permitted provided that the following conditions are
// met:
//
// * Redistributions of source code must retain the above copyright
// notice, this list of conditions and the following disclaimer.
// * Redistributions in binary form must reproduce the above
// copyright notice, this list of conditions and the following
// disclaimer in the documentation and/or other materials provided
// with the distribution.
// * Neither the name of Google Inc. nor the names of its
// contributors may be used to endorse or promote products derived
// from this software without specific prior written permission.
//
// THIS SOFTWARE IS PROVIDED BY THE COPYRIGHT HOLDERS AND CONTRIBUTORS
// "AS IS" AND ANY EXPRESS OR IMPLIED WARRANTIES, INCLUDING, BUT NOT
// LIMITED TO, THE IMPLIED WARRANTIES OF MERCHANTABILITY AND FITNESS FOR
// A PARTICULAR PURPOSE ARE DISCLAIMED. IN NO EVENT SHALL THE COPYRIGHT
// OWNER OR CONTRIBUTORS BE LIABLE FOR ANY DIRECT, INDIRECT, INCIDENTAL,
// SPECIAL, EXEMPLARY, OR CONSEQUENTIAL DAMAGES (INCLUDING, BUT NOT
// LIMITED TO, PROCUREMENT OF SUBSTITUTE GOODS OR SERVICES; LOSS OF USE,
// DATA, OR PROFITS; OR BUSINESS INTERRUPTION) HOWEVER CAUSED AND ON ANY
// THEORY OF LIABILITY, WHETHER IN CONTRACT, STRICT LIABILITY, OR TORT
// (INCLUDING NEGLIGENCE OR OTHERWISE) ARISING IN ANY WAY OUT OF THE USE
// OF THIS SOFTWARE, EVEN IF ADVISED OF THE POSSIBILITY OF SUCH DAMAGE.
// Flags: --stack-size=1200
// Create a regexp in the form of a?a?...a? so that fully
// traversing the entire graph would be prohibitively expensive.
// This should not cause time out.
var r1 = "";
for (var i = 0; i < 1000; i++) {
r1 += "a?";
}
"test".match(RegExp(r1));
var r2 = "";
for (var i = 0; i < 100; i++) {
r2 += "(a?|b?|c?|d?|e?|f?|g?)";
}
"test".match(RegExp(r2));
// Create a regexp in the form of ((..(a)a..)a.
// Compiling it causes EatsAtLeast to reach the maximum
// recursion depth possible with a given budget.
// This should not cause a stack overflow.
var r3 = "a";
for (var i = 0; i < 1000; i++) {
r3 = "(" + r3 + ")a";
}
"test".match(RegExp(r3));
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