// 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));