59a561480d
The test case examine-backward.exp issues the command "x/-s" after the end of the first string in TestStrings, but without making sure that this string is preceded by a string terminator. Thus GDB may spuriously print some random characters from before that string, and then the test fails. This patch assures that TestStrings is preceded by a string terminator. gdb/testsuite/ChangeLog: * gdb.base/examine-backward.c (Barrier): New character array constant, to ensure that TestStrings is preceded by a string terminator.
114 lines
3.9 KiB
C
114 lines
3.9 KiB
C
/* This testcase is part of GDB, the GNU debugger.
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Copyright 2015-2017 Free Software Foundation, Inc.
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This program is free software; you can redistribute it and/or modify
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it under the terms of the GNU General Public License as published by
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the Free Software Foundation; either version 3 of the License, or
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(at your option) any later version.
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This program is distributed in the hope that it will be useful,
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but WITHOUT ANY WARRANTY; without even the implied warranty of
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MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE. See the
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GNU General Public License for more details.
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You should have received a copy of the GNU General Public License
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along with this program. If not, see <http://www.gnu.org/licenses/>. */
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/*
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Define TestStrings, TestStringsH, and TestStringsW to test utf8, utf16,
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and utf32 strings respectively.
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To avoid compile errors due to old compiler mode, we don't use string
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literals. The content of each array is the same as followings:
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const char TestStrings[] = {
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"ABCD"
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"EFGHIJKLMNOPQRSTUVWXYZ\0"
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"\0"
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"\0"
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"\u307B\u3052\u307B\u3052\0"
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"012345678901234567890123456789\0"
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"!!!!!!\0"
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};
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*/
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/* This is here just to ensure we have a null character before
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TestStrings, to avoid showing garbage when we look for strings
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backwards from TestStrings. */
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const char Barrier[] = {
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0x00,
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};
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const char TestStrings[] = {
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0x41, 0x42, 0x43, 0x44, 0x45, 0x46, 0x47, 0x48,
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0x49, 0x4a, 0x4b, 0x4c, 0x4d, 0x4e, 0x4f, 0x50,
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0x51, 0x52, 0x53, 0x54, 0x55, 0x56, 0x57, 0x58,
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0x59, 0x5a, 0x00, 0x00, 0x00, 0xe3, 0x81, 0xbb,
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0xe3, 0x81, 0x92, 0xe3, 0x81, 0xbb, 0xe3, 0x81,
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0x92, 0x00, 0x30, 0x31, 0x32, 0x33, 0x34, 0x35,
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0x36, 0x37, 0x38, 0x39, 0x30, 0x31, 0x32, 0x33,
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0x34, 0x35, 0x36, 0x37, 0x38, 0x39, 0x30, 0x31,
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0x32, 0x33, 0x34, 0x35, 0x36, 0x37, 0x38, 0x39,
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0x00, 0x21, 0x21, 0x21, 0x21, 0x21, 0x21, 0x00,
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0x00
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};
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const short TestStringsH[] = {
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0x0041, 0x0042, 0x0043, 0x0044, 0x0045, 0x0046, 0x0047, 0x0048,
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0x0049, 0x004a, 0x004b, 0x004c, 0x004d, 0x004e, 0x004f, 0x0050,
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0x0051, 0x0052, 0x0053, 0x0054, 0x0055, 0x0056, 0x0057, 0x0058,
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0x0059, 0x005a, 0x0000, 0x0000, 0x0000, 0x307b, 0x3052, 0x307b,
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0x3052, 0x0000, 0x0030, 0x0031, 0x0032, 0x0033, 0x0034, 0x0035,
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0x0036, 0x0037, 0x0038, 0x0039, 0x0030, 0x0031, 0x0032, 0x0033,
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0x0034, 0x0035, 0x0036, 0x0037, 0x0038, 0x0039, 0x0030, 0x0031,
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0x0032, 0x0033, 0x0034, 0x0035, 0x0036, 0x0037, 0x0038, 0x0039,
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0x0000, 0x0021, 0x0021, 0x0021, 0x0021, 0x0021, 0x0021, 0x0000,
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0x0000
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};
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const int TestStringsW[] = {
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0x00000041, 0x00000042, 0x00000043, 0x00000044,
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0x00000045, 0x00000046, 0x00000047, 0x00000048,
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0x00000049, 0x0000004a, 0x0000004b, 0x0000004c,
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0x0000004d, 0x0000004e, 0x0000004f, 0x00000050,
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0x00000051, 0x00000052, 0x00000053, 0x00000054,
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0x00000055, 0x00000056, 0x00000057, 0x00000058,
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0x00000059, 0x0000005a, 0x00000000, 0x00000000,
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0x00000000, 0x0000307b, 0x00003052, 0x0000307b,
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0x00003052, 0x00000000, 0x00000030, 0x00000031,
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0x00000032, 0x00000033, 0x00000034, 0x00000035,
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0x00000036, 0x00000037, 0x00000038, 0x00000039,
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0x00000030, 0x00000031, 0x00000032, 0x00000033,
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0x00000034, 0x00000035, 0x00000036, 0x00000037,
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0x00000038, 0x00000039, 0x00000030, 0x00000031,
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0x00000032, 0x00000033, 0x00000034, 0x00000035,
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0x00000036, 0x00000037, 0x00000038, 0x00000039,
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0x00000000, 0x00000021, 0x00000021, 0x00000021,
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0x00000021, 0x00000021, 0x00000021, 0x00000000,
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0x00000000
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};
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int
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main (void)
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{
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/* Backward disassemble test requires at least 20 instructions in
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this function. Adding a simple bubble sort. */
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int i, j;
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int n[] = {3, 1, 4, 1, 5, 9};
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int len = sizeof (n) / sizeof (n[0]);
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for (i = 0; i < len - 1; ++i)
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{
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for (j = i; j < len; ++j)
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{
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if (n[j] < n[i])
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{
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int tmp = n[i];
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n[i] = n[j];
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n[j] = tmp;
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}
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}
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}
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return 42;
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}
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